A TEXTBOOK ON ORDINARY
DIFFERENTIAL EQUATIONS
Solutions of exercises
Shair Ahmad, Antonio Ambrosetti
ii
Contents
1 Solutions to exercises of Chapter 1
1
2 Solutions to exercises of Chapter 2
11
3 Solutions to exercises of Chapter 3
19
4 Solutions to exercises of Chapter 4
45
5 Solutions to exercises of Chapter 5
49
6 Solutions to exercises of Chapter 6
79
7 Solutions to exercises of Chapter 7
89
8 Solutions to exercises of Chapter 8
117
9 Solutions to exercises of Chapter 9
129
10 Solutions to exercises of Chapter 10
135
11 Solutions to exercises of Chapter 11
149
12 Solutions to exercises of Chapter 12
169
13 Solutions to exercises of Chapter 13
183
iii
iv
CONTENTS
Chapter 1
Solutions to exercises of
Chapter 1
1. Find the equation whose general solution is x = c e
Solution. One has x0 =
5c e
5t
. Since c e
5t
5t
.
= x we find x0 =
5x.
2. Solve x0 + (ln 3)x = 0.
Solution. An integrating factor is e(ln 3)t = 3t . Then 3t (x0 (t) +
0
(ln 3)x(t)) = (3t x(t)) = 0 and thus 3t x(t) = c whereby x(t) = c3 t .
3. Solve x0 + 4x = 4.
0
Solution. An integrating factor is e4t and the equation becomes (e4t x) =
4e4t , that is
Z
e4t x = 4
whereby x(t) = c e
4t
e4t dt = e4t + c,
+ 1.
4. Find all the solutions to the initial value problem x0 +
0, x(0) = 0.
2t3 + sin t + 5
x=
t12 + 5
Solution. x(t) ⌘ 0 is obviously one solution. Since the coefficient
functions are continuous, by the uniqueness property, this is the only
solution.
5. Solve x0 =
2x + 3 and find the solution satisfying x(1) = 5.
1
2
CHAPTER 1. SOLUTIONS TO EXERCISES OF CHAPTER 1
Solution. The general solution is x(t) = c e 2t + 32 . Setting t = 1 and
x(1) = 5, one finds 5 = c e 2 + 32 . Solving, one gets c e 2 = 5 32 = 72 ,
that is c = 72 e2 . Then the solution is x = 72 e2 · e 2t + 32 = 72 e2(1 t) + 32 .
6. Find k such that there exists a solution of x0 = kx such that x(0) = 1
and x(1) = 2.
Solution. The general solution of x0 = kx is x = c ekt . From x(0) = 1
it follows 1 = c e0 = c. From x(1) = 2 it follows 2 = c ek = ek whereby
k = ln 2. The solution becomes x = e(ln 2)t = 2t .
7. Explain why the solution to the problem
x0
2(cos t)x = cos t, x(0) =
1
2
must oscillate, i.e. it must have arbitrarily large zeros.
Solution. Using the method of integrating factors, one finds the general solution x = 12 + c e2 sin t from which we find the required solution
to be x(t) = 12 +e2 sin t . In order to show that it oscillates, we note that
x(0) = 1/2 > 0 while x( ⇡/2) = 12 + e 2 < 0. Therefore by the Intermediate Value Theorem, x(t) has a zero t0 in the interval ( ⇡/2, 0).
Since x(t) is 2⇡ periodic, the solution has arbitrarily large zeros given
by t0 + 2n⇡, n any integer number.
8. In each of the following, find the maximum interval of existence of the
solution, guaranteed by the existence theorem.
1
(a) x0 + 2
x = 0, x( 2) = 1.
t
1
1
(b) x0 + (sec t) x =
, x ( ⇡4 ) = 1.
t 1
1
Solutions. (a) The coefficient p(t) = 2
is defined and continuous
t
1
for t 6= ±1. The maximal interval that includes t0 = 2 is ( 1, 1).
In such an interval we can apply the existence theorem.
(b) The coefficient p(t) = sec t is defined and continous at t as long as
1
cos t 6= 0. The right hand side q(t) =
is defined and continuous
t 1
for t 6= 1. Thus the first point of discontinuity of the coefficients to the
left of ⇡/4 is ⇡/2 and the first such point to the right of ⇡/4 is 1.
3
Thus the maximum interval that includes t0 = ⇡/4 and where we can
apply the existence theorem is ( ⇡/2, 1).
9. Solve tx0 + x = 2t2 .
Solution. Since (tx)0 = tx0 +x we can write the equation as (tx)0 = 2t2 .
Integrating we find tx(t) = 23 t3 + c. Assuming t 6= 0 one obtains
c
x(t) = 23 t2 + .
t
10. Show that there is an infinite family of solutions to the problem
t2 x0
2tx = t5 , x(0) = 0.
all of which exist everywhere on ( 1, 1). Does this violate the uniqueness property of such equations?
Solution. Assuming t 6= 0 and solving, one obtains the family of
solutions x = 12 t4 + ct2 . All of these solutions satisfy the required
conditions. In order to apply the uniqueness theorem, we must write
2
the equation in standard form x0
x = t3 . Here the existence and
t
uniqueness theorem does not apply since the coefficient functions are
not continuous at t = 0.
11. Solve x0 = 2tx and find the solution satisfying x(0) = 4.
2
Solution. An integrating factor is e t . Then x0 2tx = 0 is equivalent
to
⇣ 2 ⌘0
2
e t (x0 2tx) = e t x = 0.
2
2
Integrating one finds e t x = c whereby x = c et . Setting t = 0 and
x(0) = 4 one finds c = 4 and hence the solution satisfying the initial
2
value is x = 4et .
12. Solve x0 =
t2 x.
Solution. An integrating factor is e
3
is x = c e t /3 .
13. Solve x0 + ax = bt.
R
t2 dt
= et
3 /3
. The general solution
4
CHAPTER 1. SOLUTIONS TO EXERCISES OF CHAPTER 1
Solution.R The equation is linear non-homogeneous. An integrating
factor is e adt = eat and hence the equation is equivalent to
Z
0
at 0
at
at
at
e (x + ax) = e x = bte
that is e x = b teat dt.
The last integral can be integrated by parts, yielding
✓
Z
Z
1 at 1
1 at
1 at 1 at
at
at
te dt = te
e dt = te
e = e
t
a
a
a
a2
a
Then, setting C = cb,
✓
◆
b at
1
at
e x= e
t
+C
a
a
b
whereby x(t) = (t
a
14. Solve: (a) x0 = x + 2t, (b) x0
2x = 3t, (c) x0 + 3x =
1
a
◆
1
) + Ce
a
+ c.
at
.
2t
Solution. We use the preceding exercise. (a) x(t) = 2(t + 1) + C et ,
(b) x(t) = 32 (t + 12 ) + Ce2t , (c) x(t) = 23 (t 13 ) + Ce 3t .
15. Find the solution of x0 + ax = bt satisfying x(t0 ) = x0 .
Solution. The general solution is
✓
b
x(t) =
t
a
1
a
◆
at
+ ce
.
To find c, we put t = t0 and x(t0 ) = x0 . One finds
✓
◆
b
1
x0 =
t0
+ ce at0
a
a
that is
c=e
at0
✓
x0
b
a
✓
and hence the solution is
✓
◆
✓
b
1
at0
x(t) =
t
+e
x0
a
a
1
a
t0
b
a
✓
◆◆
t0
1
a
◆◆
e
at
.
5
16. Solve the initial value problems (a) x0 x = 12 t, x(0) = 1, (b) x0 +x = 4t,
x(1) = 0, (c) x0 2x = 2t, x(0) = 3.
Solution. Using the preceding exercise we find (a) x(t) =
3 t
e , (b) x(t) = 4(t 1), (c) x(t) =
t + 12 + 52 e2t .
2
1
(t + 1) +
2
17. Given h, k 2 R, k > 0, find the limits as t ! +1 of the solutions of
x0 + kx = h.
Solution. The general solution is xc (t) = ce kt + hk and since k > 0
we find
✓
◆
h
h
kt
lim xc (t) = lim ce +
= .
t!+1
t!+1
k
k
18. Consider x0 + kx = 1, where k is a constant.
(a) For what value of k will all solutions tend to 2 as t ! +1?
(b) Is there any value of k for which there exists a non-constant solution
x(t) such that x(t) ! 3 as t ! +1? Explain.
Solutions. (a) We find that the general solution is x = 1/k + ce kt . It
is easy to see that we must have k > 0. Hence ce kt approaches 0 as t
approaches 1, and all solutions will approach 2 if and only if 1/k = 2,
or k = 1/2.
(b) From (a) k has to be = 1/3. Then x = 3 + cet/3 with c 6= 0,
otherwise x is constant. It is clear that limt!+1 x(t) is either +1, if
c > 0, or 1, if c < 0.
19. Find the limits as t ! ±1 of the solution of x0 =
1
1+t2
x, x(0) = 1.
Solution. An integrating factor is
e
R
1
dt
1+t2
=e
arctan t
and hence the general solution is x(t) = c earctan t . Here and below
arctan t denotes the principal part of the arctangent function whose
image is ( ⇡/2, ⇡/2). The initial condition x(0) = 1 yields c = 1 and
hence x(t) = earctan t . Since limt!±1 (arctan t) = ±⇡/2, then
lim earctan t = e
t! 1
⇡/2
,
lim earctan t = e⇡/2 .
t!+1
6
CHAPTER 1. SOLUTIONS TO EXERCISES OF CHAPTER 1
20. Consider x0 + kx = h, with k 6= 0. Find conditions on the constants
h, k such that
(a) all solutions tend to 0 as t tends to + infinity.
(b) it will have only one solution bounded on (0, +1).
(c) all solutions are asymptotic to the line x = 3.
Solution. Solving the equation, we find that the gereral solution is
xc = h/k + ce kt . So, it follows that :
(a) xc (t) ! 0 as t ! +1 for all c, if and only if h = 0 and k > 0.
(b) If c 6= 0 and k > 0, then every solution xc is bounded on [0.1). If
c 6= 0 and k < 0, then every solution cc is unbounded. Therefore, only
for c = 0 there is a unique bounded solution
h
(c) If k > 0 and h = 3k then for every c one has xc (t) ! = 3.
k
21. Show that for any di↵erentiable function f (t), t 2 R, all solutions of
x0 + x = f (t) + f 0 (t) tend to f (t) as t tends to +1.
Solution. Solving for x, we have
Z
t
e x = et (f (t) + f 0 (t))dt + c
and hence
x(t) = e
e
t
Z
t
Z
et (f (t) + f 0 (t))dt + ce
t
e f (t)dt + e
t
Z
t
=
et f 0 (t)dt + ce t .
Integrating the second integral by parts, we obtain x(t) = f (t) + ce t ,
which approaches f (t) as t approaches 1.
22. Find a continuous function q(t), t 2 R, such that all the solutions of
x0 + x = q(t)
(a) approach the line x = 7t
(b) approach the curve x = t2
5 as t ! +1.
2t + 5 as t ! +1.
7
Solution. To answer to (a) we take f (t) = 7t 5, which gives f (t) +
f 0 (t) = 7t + 2 and therefore all solutions of x0 + x = 7t + 2 tend to
7t 5.
As for (b) we take f (t) = t2 2t + 5, which gives f (t) + f 0 (t) = t2 + 3
Thus all solutions of x0 + x = t2 + 3 tend to t2 2t + 5.
23. Show that if p is di↵erentiable and such that limt!+1 p(t) = +1 then
all the solutions of x0 + p0 (t)x = 0 tend to zero as t ! +1.
R
0
Solution. An integrating factor is e p (t)dt = ep(t) and the general
solution is x(t) = c e p(t) . If limt!+1 p(t) = +1 then
p(t)
lim c e
t!+1
= ce
limt!+1 p(t)
= 0.
24. If k 6= 0, show that the constant solution x(t) = k12 is the only solution
of x0 k 2 x = 1 such that the limt!+1 x(t) is finite.
Solution. An integrating factor is e
k2 t
(e
k2 t
. Then
x(t))0 = e
k2 t
which yields
1 k2 t
e
+c
k2
2
and hence x(t) = k12 + cek t . It follows that x(t) has a finite limit as
1
t ! +1 if and only if c = 0 and this yields x(t) =
.
k2
e
k2 t
x(t) =
25. Let kR 6= 0 and let q(t) be continuous and such that limt!+1 q(t) = 0
2
+1
and 0 e k s q(s)ds = 0. Show that the solution x of the ivp problem
x0
k 2 x = q(t),
x(0) = x0 ,
tends to 0 as t ! +1 if and only if x0 = 0.
Solution. The solution of x0 k 2 x = q(t), x(0) = x0 , is

Z t
Z t
2
2
k2 t
k2 s
k2 t
x(t) = e
x0 +
e
q(s)ds = e
e k s q(s)ds + x0 ek t .
0
Letting
(t) = e
k2 t
0
Z
t
e
0
k2 s
q(s)ds =
Rt
0
e
k2 s
e
q(s)ds
k2 t
,
8
CHAPTER 1. SOLUTIONS TO EXERCISES OF CHAPTER 1
2
we can write x(t) as x(t) = (t) + x0 ek t . To find the limit of x(t) as
t ! +1 let us first evaluate
!
R t k2 s
e
q(s)ds
0
lim (t) = lim
.
t!+1
t!+1
e k2 t
From limt!+1 q(t) = 0 it
in particular,
q is bounded on
R t follows,
R +1 kthat
2s
k2 s
[0, +1). Then limt!+1 0 e
q(s)ds = 0 e
q(s)ds = 0, and the
0
right hand side is the undetermined form 0 . Using the Hopital rule one
finds
!
!
R t k2 s
k2 t
e
q(s)ds
e
q(t)
q(t)
0
lim
= lim
= lim
= 0.
2t
2t
k
2
k
t!+1
t!+1
t!+1
k2
e
k e
2
We have shown that x(t) = (t) + x0 ek t with limt!+1 (t) = 0. Then
x0 > 0 ) lim x(t) = +1,
t!+1
x0 < 0 ) lim x(t) =
t!+1
1
while if x0 = 0 we find x(t) = (t) ! 0.
26. Show that the solution of x0 = k 2 x, x(t0 ) = x0 , is increasing if x0 > 0
and decreasing if x0 < 0.
Solution. A first method is to use directly the equation, without
solving it. We know that the nontrivial solutions of x0 = k 2 x do not
change sign. If x(t0 ) = x0 > 0 then x0 (t) remains positive for t
t0
and hence the solution x(t) is increasing. Similarly, if x0 < 0, then x(t)
is decreasing.
2
Another way is to solve the equation, yielding x = c ek t . From x(t0 ) =
2
2
x0 it follows that x0 = c ek t0 whereby c = x0 e k t0 . Thus
x(t) = x0 e
k 2 t0 k 2 t0
e
= x0 e k
2 (t
t0 )
.
It follows that x(t) is strictly increasing if x0 > 0 whilst x(t) is strictly
decreasing if x0 < 0.
27. Show that the solution of x0 = kx, x(t0 ) = x0 is increasing if kx0 > 0
and decreasing if kx0 < 0.
Solution. We can solve the ivp finding x(t) = x0 ek(t t0 ) . Since x0 = kx,
x0 > 0 if and only if kx > 0. As before the sign of x is the same as the
sign of x0 . So x0 = kx > 0 if and only if kx0 > 0.
9
28. Find the locus of minima of the solutions of x0 + 2x = 6t.
Solution. At the points where the minima occur, x0 = 0 and hence
2x = 6t, that is x = 3t. Moreover, from x0 + 2x = 6t, di↵erentiating,
we infer x00 + 2x0 = 6. On the straight line x = 3t we have x0 = 0 and
hence x00 = 6 > 0. Then these points are minima.
x
O
t
x=3t
Figure 1.1: Solutions of x0 + 2x = 6t for c > 0 (in black), for c < 0 (in red)
and for c = 0 (in blue). The dotted line x = 3t is the locus of minima.
We might also solve the equation finding that the general solution is
1
x(t) = 3(t
) + c e 2t . The reader can check that this family of
2
functions have their minima on x = 3t provided c > 0. If c  0 the
functions are strictly increasing and do not have minima. See figure
1.1.
29. Find the locus of maxima and minima of the solutions of x0 + x = at,
a 6= 0.
Solution. At the points where the minima occur, x0 = 0 and hence
x = at. From x00 + x0 = a and x0 = 0 we infer x00 = a. Hence if a > 0
10
CHAPTER 1. SOLUTIONS TO EXERCISES OF CHAPTER 1
the straight line x = at is the locus of minima, while if a < 0 it is the
locus of maxima.
Solving the equation we find x = ce t +at a. Notice that x0 = ce t +a
and thus if a > 0 and c < 0 one has x0 > 0, the corresponding solutions
are increasing. Similarly, if a < 0 and c > 0 the solutions are decreasing.
figure 1.2.
x
x=at
x=at
x
t
(a)
t
(b)
Figure 1.2: (a) a > 0. In red the solutions with c < 0. (b) a < 0. In red the
solutions with c > 0
Chapter 2
Solutions to exercises of
Chapter 2
1. Check that the local existence and uniqueness theorem applies to the
ivp x0 = t + x2 , x(0) = 0.
Solution. The function f (t, x) = t + x2 is continous and continuously
di↵erentiable with respect to x..
2. Show that the function f (x) = |x|p is not lipschitzian at x = 0 if
0 < p < 1.
Solution. If f is lipschitzian at x = 0, there exist ✏ > 0 and L > 0
such that ||x|p |y|p |  L|x y| for all x, y 2 ( ✏, ✏). In particular,
taking y = 0, we get |x|p  L|x| for all x 2 ( ✏, ✏), which is false
because 0 < p < 1.
3. Find a such that the existence and uniqueness theorem applies to the
ivp x0 = 32 |x|1/3 , x(0) = a.
Solution. The function f (x) = |x|1/3 is infinitely many times di↵erentiable for all x 6= 0. Thus for all a 6= 0 the existence and uniqueness
theorem applies. If a = 0 the function f (x) = |x|1/3 is continuous
but not lipschitzian. By inspection, one checks that, in addition to
x(t) ⌘ 0, the function x = t3/2 , t > 0, solves the ivp.
4. Check that for all t0 , a 2 R the existence and uniqueness theorem
applies to the ivp ln x0 = x2 , x(t0 ) = a.
11
12
CHAPTER 2. SOLUTIONS TO EXERCISES OF CHAPTER 2
2
Solution. Solving for x0 , the equation can be written as x0 = ex . The
2
function f (x) = ex is infinitely many times di↵erentiable for all x.
t
5. Explain why x0 + esin
t +1 x = 0 cannot have a solution x(t) such that
x(1) = 1 and x(2) = 1.
Solution. By uniqueness, any nontrivial solution of the equation cannot vanish. On the other hand, by the Intermediate Value theorem,
there is t⇤ 2 (1, 2) such that x(t⇤ ) = 0.
0
6. Transform the equation ex = x into an equation in normal form and
show that it has a unique solution such that x(t0 ) = a, for all t0 and
all a > 0.
Solution. One has x0 = ln x (x > 0). The function f (x) = ln x is
infinitely many times di↵erentiable for all x > 0 and hence the existence
and uniqueness theorem applies, yielding a unique solution such that
x(t0 ) = a > 0.
7. Find the equation whose solution is the catenary x(t) = cosh t = 12 (et +
e t ).
Solution. x0 = 12 (et e t )(= sinh t). Then x02 = 14 (et
1 2t
1
(e + e 2t 2) = 14 (e2t + e 2t )
= 1 + 14 (e2t + e
4
2
1 2t
1 t
2t
t 2
1 + 4 (e + e + 2) = 1 + 4 (e + e ) = 1 + x2 .
8. Check that the functions x(t) ⌘ 1 and
⇢
sin t if ⇡2  t 
(t) =
1
if t > ⇡2
are solutions of the ivp x0 =
p
1
e t )2 =
2t
) + 12 =
⇡
2
x2 , x( ⇡2 ) = 1.
Solution. Clearly x ⌘ 1 solves the ivp. As for , first of all it is
di↵erentiable. This is trivial for t 6= ⇡2 . Moreover, for t 6= ⇡2 one has
0
(t) =
⇢
cos t if ⇡2  t <
0
if t > ⇡2
⇡
2
Thus
lim
t! ⇡2
0
(t) = lim
cos t = 0 = lim
⇡
⇡
t! 2
t! 2 +
0
(t).
13
This shows that (t) is di↵erentiable at t = ⇡2 , also. Let us check that
solves the ivp. This is trivial for t > ⇡2 . If ⇡2  t  ⇡2 , one has
p
0
that
(t)
=
sin
t
satisfies
(t)
=
cos
t
>
0.
Then
cos
t
=
1 sin2 t =
p
2 (t), as well as ( ⇡ ) = sin ⇡ = 1.
1
2
2
Notice that we can define (t) ⌘
⇡
.
2
1 for t 
9. Find a 0 such that the Cauchy problem x0 = |x|1/4 , x(0) = a has a
unique solution.
Solution. The function f (x) = |x|1/4 is lipschizian for x > 0. Actually
it is continuously di↵erentiable with f 0 (x) = 14 x 3/4 , x > 0. Thus if
a > 0 the ivp has a unique solution. If a = 0, x(t) ⌘ 0 is a solution,
together with all the functions
8
for t  c
< 0
4/3
xc (t) =
3
:
(t c)
for t > c.
4
10. Show that if p > 1 the solution of x0 = xp , x(0) = a > 0 is not defined
for all t 0.
Solution. The solution is positive and can be found integrating x p x0 =
1. For t 0 one has
Z
t
x p x0 dt = t
0
and then
x(t)1 p a1
1 p
p
= t,
ore equivalently a1
p
x(t)1
p
= (p
1)t,
whereby
x(t) = a1
p
(p
Since p > 1, x(t) is defined for a
1
.
(p 1)ap 1
1 p
1)t
(p
1/(p 1)
.
1)t > 0, namely for t <
11. Show that if 0 < p  1, the solution of x0 = |x|p , x(0) = a > 0 is
defined for all t 0.
Solution. We can argue as in the preceding exercise noticing that now
the exponent 1/(p 1) is positive.
14
CHAPTER 2. SOLUTIONS TO EXERCISES OF CHAPTER 2
12. Show that the solutions of x0 = sin x are defined on all t 2 R.
Solution. The function f (x) = sin x is continuously di↵erentiable and
f 0 (x) = cos x is bounded. Then we can apply the Global Existence
Theorem.
13. Show that the solutions of x0 = arctan x are defined on all t 2 R.
Solution. f (x) = arctan x is continuously di↵erentiable and f 0 (x) =
1/(1 + x2 ) is bounded.
14. Show that the solutions of x0 = ln(1 + x2 ) are defined on all t 2 R.
Solution. The function f (x) = ln(1+x2 ) is continuously di↵erentiable
and f 0 (x) = 2x/(1+x2 ) ! 0 is bounded. Thus we can once again apply
the Global Existence Theorem.
15. Show that the ivp x0 = max{1, x}, x(0) = 1, has a unique solution
defined for all t and find it.
Solution. The function f (x) = max{1, x} is globally lipschitzian (but
not di↵erentiable at x = 0) and hence the ivp has a unique solution
defined for all t. Since x0
1, the solution is strictly increasing. For
t > 0 one has x(t) > x(0) = 1 and hence f (x) = x. Then x0 = x
which yields x(t) = cet . Using the initial condition, we get c = 1. For
t < 0 one has x(t) < 1 and hence f (x) = 1. Then x0 = 1 which implies
x = t + c. The initial condition yields c = 1. In conclusion the solution
is given by
⇢
et
for t 0
x(t) =
t + 1 for t  0.
Notice that this function is continuously di↵erentiable for all t (but it
is not twice di↵erentiable). Actually, x0 (t) = 1 for t < 0 and x0 (t) = et
for t > 0. Moreover, taking the limits of the left and right incremental ratios one has limt!0 x(t) t x(0) = limt!0 t+1t 1 = 1 as well as
t
0
t
limt!0+ x(t) t x(0) = limt!0+ e t e = limt!0+ e t 1 = (using the Hopital
rule) = limt!0+ et = 1. This shows that x is di↵erentiable also at t = 0
and x0 (0) = 1. Clealry x0 is continuous.
16. Show that the ivp x0 = max{1, x}, x(0) =
defined for all t and find it.
1, has a unique solution
15
x
O
t
Figure 2.1: Solution of x0 = max{1, x}, x(0) = 1
Solution. As before, the function f (x) = max{1, x} is globally lipschitzian. One finds that x(t) x(0) = 1 for t 0 and hence x0 (t) = 1
for t 0. For t < 0 one has x(t)  1 and hence x0 (t) = x. It follows
that the solution is
⇢
t 1 for t 0
x(t) =
et for t  0.
17. Show that the solution of x0 = t2 x4 + 1, x(0) = 0 is odd.
Solution. The function z(t) = x( t) solves z 0 = x0 ( t) = ( t)2 x4 ( t)+
1 = t2 ( z)4 + 1 = t2 z 4 + 1, as well as z(0) = x(0) = 0. By uniqueness,
x(t) = z(t), that is x(t) = x( t).
18. Show that, if f (x) > 0, resp. f (x) < 0, the solutions of x0 = f (x)
cannot be even.
Solution. Since x0 = f (x) > 0, resp. < 0, all the solutions are strictly
increasing, resp. decreasing, and hence cannot be even.
19. Show that the solution x(t) of the Cauchy problem x0 = 2 + sin x,
x(0) = 0, cannot vanish for t > 0.
Solution. The solution is defined for all t. Since sin x
x0 = 2 + sin x > 1. Therefore x(t) > x(0) = 0 for all t > 0.
1 then
20. Let f be continuously di↵erentiable and such that f (0) = 0. Show that
the solutions of x0 = f (x)h(t) cannot change sign.
16
CHAPTER 2. SOLUTIONS TO EXERCISES OF CHAPTER 2
Solution. Since f (0) = 0, x(t) ⌘ 0 is a solution of the equation. If x(t)
changes sign, then there exists t0 such that x(t0 ) = 0. By uniqueness,
the solution of the Cauchy problem x0 = f (x)h(t), x(t0 ) = 0 is x(t) ⌘ 0.
21. Show that the solutions of x0 = sin(tx) are even.
Solution. f (t, x) = sin(tx) is odd with respect to t because sin( tx) =
sin(tx). Using Proposition 2.3.3 the conclusion follows.
22. Find the limits, as t ! ±1, of the solution
(x + 2)(1 x4 ), x(0) = 0.
(t) of the ivp x0 =
Solution. Letting f (x) = (x + 2)(1 x4 ), one has that f (±1) = 0 and
f (x) > 0 for 1 < x < 1. Then (t) is increasing and limt! 1 (t) =
1, limt!+1 (t) = +1.
23. Show that for every 1 < a < 1 the solution (t) of the ivp x0 = x3
x(0) = a is such that limt!+1 (t) = 0.
x,
Solution. Letting f (x) = x3 x one has that f (0) = f (±1) = 0,
f (x) > 0 for x 2 ( 1, 0) and f (x) < 0 for x 2 (0, 1). If a 2 ( 1, 0)
then (t) is increasing and limt!+1 (t) = 0. If 0 < a < 1 then (t) is
decreasing and limt!+1 (t) = 0 as well. Finally if a = 0 then (t) ⌘ 0.
24. Show that the solutions of x0 = arctan x + t cannot have maxima.
Solution. Setting f (t, x) = arctan x + t, let t⇤ be such that x0 (t⇤ ) = 0.
Then x00 (t⇤ ) = ft (t⇤ , x(t⇤ )) = 1 > 0. Thus at t⇤ the solution x cannot
have a maximum.
25. Show that the solutions of x0 = ex
t cannot have minima.
Solution. Setting f (x) = ex t, let t⇤ be such that x0 (t⇤ ) = 0. Then
x00 (t⇤ ) = ft (t⇤ , (t⇤ )) = 1. Thus at t⇤ the solution x cannot have a
minimum.
26. Let (t) be the solution of the ivp x0 = tx
Show that has a maximum at t = a.
t3 , x(a) = a2 with a 6= 0.
Solution. One has that 0 (a) = a (a) a3 = 0. Moreover, from
00
(t) = (t)+t 0 (t) 3t2 it follows that 00 (a) = (a) 3a2 = a2 3a2 =
2a2 < 0.
17
27. Let (t) be the solution of the ivp x0 = tx
Show that has a minimum at t = 0.
Solution. One has that
t 0 (t) 3t2 it follows that
0
t3 , x(0) = a2 with a 6= 0.
(0) = 0. Moreover, from
(0) = (0) = a2 > 0.
00
(t) =
(t) +
00
28. Show that the solution (t) of x0 = x2
point at t = 0.
t2 , x(0) = 0, has an inflection
Solution. One has that 0 (0) = 0. Moreover, 00 (t) = 2 (t) 0 (t) 2t
and 000 (t) = 2 02 (t)+2 (t) 00 (t) 2 whereby 00 (0) = 2 (0) 0 (0) 2·0 =
0 and 000 (0) = 2 02 (0) + 2 (0) 00 (0) 2 = 2.
29. Suppose that g(x) is continuously di↵erentiable and let (t) be the
solution of x0 = tg(x), x(0) = a. If g(a) > 0, show that the function
(t) has a minimum at t = 0, for all a 2 R.
Solution. One has that 0 (0) = 0. Moreover, from 00 (t) = g( (t)) +
tg 0 ( (t)) 0 (t) it follows that 00 (0) = g( (0)) = g(a) > 0.
30. Show that the solution xa (t) of x0 = 2t + g(x), xa (0) = a > 0 satisfies
xa (t) t + t2 for t > 0, provided g(x) 1.
Solution. The solution of y 0 = 1 + 2t such that y(0) = 0 is y = t2 + t.
If g(x) 1 then, by comparison, xa (t) t + t2 .
31. Let xa (t) be the solution of of x0 = t + g(x), xa (0) = a,with 0 < a <
2. If xa (t) is defined for all t > 0 and g(x)  x, show, using the
comparison theorem 2.3.6, that the equation xa (t) = 0 has at least one
positive solution in (0, 2).
Solution. The solution of y 0 = t y such that y(0) = 2 is y =
e t t + 1. If g(x)  x then, by comparison, xa (t)  e t t + 1 for
t > 0. Thus xa (2) < 0. Since xa (0) = a > 0 then by the intermediate
value theorem there exists t0 2 (0, 2) such that xa (t0 ) = 0.
18
CHAPTER 2. SOLUTIONS TO EXERCISES OF CHAPTER 2
Chapter 3
Solutions to exercises of
Chapter 3
1. Solve x0 = x2 + 1.
Solution. Dividing by x2 + 1 > 0 and integrating we get
Z
x0
=t+c
x2 + 1
wich yields arctan x = t + c. Solving for x we find x = tan(t + c), with
⇡
< t + c < ⇡2 .
2
2. Solve the ivp x0 = x2
1, x(0) = 0.
Solution. The equation has two constant solutions given by x = ±1.
All the other solutions are such that x(t) 6= ±1 for all t. To find the
x0
non constant solutions we integrate 2
= 1 yielding
x
1
Z
x0
= t + c.
x2 1
Since
1
x2
we find
Z
x0
x2
1
=
1
2
1
=
1
2
✓Z
✓
x0
x
1
19
1
x
1
Z
1
x+1
x0
x+1
◆
◆
=
1
x 1
ln
.
2
x+1
20
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
Then
1
ln
2
x 1
x+1
x 1
= e2(t+c) = ke2t , k = e2c . If
x+1
x 1
1 for all t then
> 0 and thus
x+1
= t + c, whereby
x(t) > 1 or x(t) <
x 1
= ke2t .
x+1
If
1 < x(t) < 1 for all t then
(i)
x 1
< 0 and thus
x+1
x 1
=
x+1
ke2t .
(ii)
Solving for x we find
1 + ke2t
1 ke2t
1 ke2t
x=
1 + ke2t
x=
in the case (i)
in the case (ii)
To find the solution such that x(0) = 0 we first notice that we are in
x(0) 1
the case (ii). Then from
= k we deduce k = 1. Therefore
x(0) + 1
1 e2t
x=
.
1 + e2t
3. Solve the ivp x0 = x2 + x, x(1) = 1
Solution. This is a kind of logistic equation x0 = x(↵
x) with
↵ = 1, = 1. Repeating the arguments carried out in Section 3.1.1,
we find that x = 0 and x = 1 are the two constant solutions. All the
other solutions satisfy one of the following alternatives
(i) x(t) > 0 for all t,
(ii)
1 < x(t) < 0 for all t,
(iii) x(t) <
1 for all t.
In particular, from x(1) = 1 we deduce that the solution is positive.
x0
To find the non constant solutions we integrate 2
= 1 yielding
x +x
Z
x0
= t + c.
x2 + x
21
Since
x2
one has
Z
x0
=
x2 + x
Z
dx
x
Z
1
1
=
+x
x
1
x+1
dx
= ln |x|
x+1
ln |x + 1| = ln
x
.
x+1
x
= t + c. Moreover, recalling that x > 0 we infer that
x+1
x
x
> 0 and hence
= et+c = ket , k = ec . From x(1) = 1 we infer
x+1
x+1
1
1
x
1
1
= ke, whereby k =
and thus
= et = et 1 . Solving for
2
2e
x+1
2e
2
x it follows
1 t 1
e
et 1
x= 2 1 t 1 =
2 et 1
1 2e
Thus ln
x2 + x
, x(0) = 1.
2x + 1
2x + 1 0
Solution. Integrating 2
x = 1 we find ln |x2 + x| = t + c. The
x +x
initial conditions yield c = ln 2. Thus ln |x2 + x| = t + ln 2, whereby
|x2 + x| = et+ln 2 = 2et . Notice that, by uniqueness, the solution is
positive. Then x2 + x = 2et and solving we find
p
1 + 1 + 8et
x=
.
2
4. Solve the ivp x0 =
5. Let
limt!
(t) be the solution of the ivp x0 =
1
(t) and limt!+1 (t).
x2
2x
x
, x(0) = 2. Find
1
Solution. First of all, we notice that x = 0 and x = 1 are the constant
solutions. By uniqueness, all the non constant solutions cannot cross
x = 0 and x = 1. Taking into account the initial condition x(0) = 2 we
infer that (t) > 1 for all t. Arguing as in the previous problem one
finds that the general solution of the equation is
x2
x = ec et .
22
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
The initial condition x(0) = 2 yelds 2 = ec . Thus (t) satisfies
| 2 (t)
(t)| = 2et . Recalling that (t) > 1 we find 2 (t)
(t) = 2et .
Then
p
1 + 1 + 8et
(t) =
2
whereby
p
1 + 1 + 8et
lim (t) = lim
=1
t! 1
t! 1
2
and
lim
t!+1
(t) = lim
1+
p
1 + 8et
= +1.
2
t!+1
6. Solve x0 = 4t3 x4 .
Solution. Either x(t) ⌘ 0 or from x 4 x0 = 4t3 it follows
1
= t4 + c
3x3
and thus, for t4 + c 6= 0,
x3 (t) =
7. Solve x0 =
1
.
+ c)
3(t4
tx2 .
Solution. Either x(t) ⌘ 0 or x 2 x0 = t. Then x 1 =
hence
p
1
2
x(t) = 1 2
= 2
, t 6= ± 2c.
t
2c
t
c
2
1 2
t
2
+ c and
8. Solve x0 = et (1 + x2 ), x(0) = 0.
x0
= et we find tan x = et + c. The initial
1 + x2
condition yields c = 1. Then arctan x = et 1 whereby x = tan(et 1),
with ⇡2 < et 1 < ⇡2 , namely 1 ⇡2 < et < 1 + ⇡2 . Since 1 ⇡2 < 0 the
solution is defined for all t < ln(1 + ⇡2 ).
Solution. Integrating
9. Solve x0 =
p
t
and find the solution such that x( 2) = 1.
x
23
Solution. Integrating xx0 = t it follows 12 x2 = 12 t2 + c which is the
p
general solution in implicit form. Setting t = 2 and x =p1 we find
c = 12 . Thus the solution satisfying the initial condition x( 2) = 1 is
1 2
x = 12 t2 12 , namely x2 = t2 1. Solving and taking into account that
2
p
the solution has to be positive, we find x = t2 1. This function is
defined for t 2 ( 1, 1] [ [1, +1) but, as a solution of our di↵erential
equation, we have to take t 2 [1, +1).
t
and find the solution such that x(1) = 1.
4x3
Solution. Integrating 4x3 x0 = t we get
10. Solve x0 =
x4 =
1 2
t + c.
2
which is the general integral in implicit form. Putting t = 1 and x = 1
we find c = 1 + 12 = 32 . Thus the solution (which is positive) of the ivp
is
✓
◆1/4 ✓
◆1/4
3 1 2
3 t2
x=
t
=
,
2 2
2
p
p
defined for
3  t  3.
11. Solve x0 =
t2 x2 , such that x(1) = 2.
Solution. The trivial solution x(t) ⌘ 0 does not satisfy the initial
condition x(1) = 2. Then, by uniqueness, x(t) > 0. From x 2 x0 = t2
we infer
1
=
x(t)
t3
+c
3
=)
x(t) =
1
t3
3
c
=
3
t3
3c
.
Putting in the preceding equation t = 1 and x(1) = 2 one gets 2 =
3/(1 3c) and hence c = 1/6. Thus the solution is given by
x(t) =
p
12. Solve x0 = 5t x, x
t3
3
+
1
2
=
2t3
6
,
+1
t>
1
21/3
.
0, x(0) = 1.
Solution. Note that x0 (t) > 0 for t > 0 and x0 (t) < 0 for t < 0. Thus
any solution has a minimum at t = 0 and hence x(t) x(0) = 1 for all
24
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
p
t. Moreover, notice that the function x is lipschitzian for x 1. This
allows us to use the arguments carried out in the preceding exercises.
Then integrating x 1/2 x0 = 5t we infer
2
p
p
5
5
1
x(t) = t2 + c i.e.
x(t) = t2 + c.
2
4
2
1
From x(0) = 1 it follows 1 = c, namely c = 2. Thus the solution is
2
✓
◆2
p
5 2
5 2
x(t) = t + 1, i.e. x(t) =
t +1 .
4
4
p
13. Solve x0 = 4t3 x, x
0, x(0) = 1.
Solution. Arguing as in the preceding exercise, any solution x(t) is
such that x(t)
x(0) = 1. Then integrating x 1/2 x0 = 4t3 it follows
that
p
2 x(t) = t4 + c.
Putting t = 0, x(0) = 1 one finds 2 = c. Then the solution of the initial
value problem is
p
2
2 x(t) = t4 + 2, i.e. x(t) = 12 t4 + 1 .
p
14. Solve and discuss uniqueness for the ivp x0 = 2t x, x(a) = 0, x
Solution. We have
2
p
0.
x(t) = t2 + c
Putting t = a and x(a) = 0, we find 0 = a2 + c, that is c =
solution becomes
2
x(t) = 12 t2 12 a2 .
a2 . The
In addition, also x(t) ⌘p0 solves the equation and the initial condition
x(a) = 0. Notice that x is not lipschitzian at x = 0 and we cannot
expect to have uniqueness.
15. Find p such that the solutions of x0 =
(p+1)tp x2 tend to 0 as t ! +1.
Solution. Let p + 1 > 0. Integrating the separable equation we find
x 1 (t) = tp+1 + c,
i.e. x(t) =
1
,
+c
tp+1
t 6=
c1/(p+1)
25
Since p + 1 > 0 then tp+1 + c ! +1 as t ! +1 and thus
1
= 0.
+c
lim x(t) = lim
t!+1 tp+1
t!+1
16. Find the limit as t ! +1 of the solutions of x0 =
p + 1  0.
(p + 1)tp x2 when
Solution. If p + 1 < 0 then tp+1 + c ! c as t ! +1 and hence
lim
1
1
= ,
+c
c
t!+1 tp+1
provided c 6= 0. If c = 0,
lim
1
t!+1 tp+1
= +1.
The case p + 1 = 0 leads to x0 = 0 and hence x(t) is a constant, which
means that its limit is the same.
p
17. Solve x0 = 1 x2 and find the singular solutions. Explain why
uniqueness does not hold.
p
Solution. The equation is separable with h(t) = 1, g(x) = 1 x2 .
Solving, we have
arcsin x(t) = t + c,
or
x(t) = sin(t + c),
c 2 R,
⇡
⇡
t+c .
2
2
The restriction on t cannot be eliminated because outside the interval
1
⇡  t + c  12 ⇡ the sine function has negative derivative, while from
2
the equation it follows the x0 (t) 0. In addition, x(t) ⌘ ±1 solve the
given equation. They are the envelope of the family x = sin(t + c) and
hence they are the singular solutions. For each initial condition
x(t0 ) =
p
±1 we do not have uniqueness. Notice that the function 1 x2 is not
lipschitzian at x = ±1 and hence we cannot apply the Existence and
Uniqueness Theorem.
26
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
x
1
O
t
-1
Figure 3.1: Solutions of x0 =
18. Solve (2x2 + 1)dx = (y 5
p
1
x2
1)dy.
Solution. The equation is both separable and exact. For example,
since it is separable, we can integrate both sides obtaining
✓ 6
◆
2 3
y
x +x=
y + c,
c a constant.
3
6
The same result could be found taking the antiderivative of ! = (2x2 +
1)dx (y 5 1)dy.
19. Solve (x + 3y)dx + (3x + y)dy = 0 and sketch a graph of the solutions.
Solution. The equation is exact and the general solution is
Z x
Z y
1
1
sds +
(3x + s)ds = x2 + 3xy + y 2 = c.
2
2
0
0
If c 6= 0 they are hyperbolas. If c = 0 then x2 + 6xy + y 2 = 0 reduces
to the pair of straight lines
p
p
y = ( 3 2 2)x,
y = ( 3 + 2 2)x,
that are the asymptotes to the hyperbolas. See fig.3.2.
27
y
x
Figure 3.2: x2 + 6xy + y 2 = 2c, c > 0 in red, c < 0 in blue, c = 0 in black
20. Solve (x + y)dx + (x
y)dy = 0.
Solution. The equation is exact and the general
Z x
Z y
1
sds +
(x s)ds = x2 + xy
2
0
0
solution is
1 2
y = c.
2
21. Solve (axp + by)dx + (bx + dy q )dy = 0, p, q > 0.
Solution. Since
@
@
(axp + by) = b =
(bx + dy q )
@y
@x
the equation is exact. The general solution is
Z x
Z y
axp+1
dy q+1
p
F (x, y) =
as ds +
(bx + dsq )ds =
+ bxy +
= c.
p+1
q+1
0
0
22. Solve (3x2 y)dx + (4y 3
through (1, 1).
x)dy = 0 and find the solution passing
Solution. The equation is exact and the general solution is
Z x
Z y
2
F (x, y) =
3s ds +
(4s3 x)ds = x3 xy + y 4 = c.
0
0
28
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
For x = y = 1 we find c = 1 and hence the solution passing through
(1, 1) is x3 xy + y 4 = 1.
23. Solve (y
(0, 1).
x1/3 )dx+(x+y)dy = 0 and find the solution passing through
Solution. Since
@
(y
@y
x1/3 ) = 1 =
@
(x + y)
@x
the equation is exact. The general solution is
3 4/3
x
4
F (x, y) =
+ xy + 12 y 2 = c.
For x = 0, y = 1 one has that c = 12 . Thus the solution passing through
(0, 1) is given by 34 x4/3 + xy + 12 y 2 = 12 .
1 2
24. Solve (ex
y )dx + (ey
2
through (0, 0).
xy)dy = 0 and find the solution passing
Solution. Since
@ x
(e
@y
1 2
y )
2
=
y=
@ y
(e
@x
xy)
the equation is exact. An antiderivative of ! = (ex 12 y 2 )dx+(ey xy)dy
is
Z x
Z y
s
F (x, y) =
e ds +
(es xs)ds = ex + ey 12 xy 2 2.
0
0
x
Thus the general solution is e +ey 12 xy 2 = c, c constant. For x = y = 0
one has that c = 2. Thus the solution passing through (0, 0) is given
by ex + ey 12 xy 2 = 2.
25. Solve (x+sin y)dx+x cos ydy = 0 and find the solution passing through
(2, ⇡).
Solution. Since
@
@
(x + sin y) = cos y =
(x cos y)
@y
@x
the equation is exact. The general solution is
Z x
Z y
sds +
x cos sds = 12 x2 + x sin y = c.
0
0
29
For x = 2, y = ⇡ one has that c = 2. Thus the solution passing through
(2, ⇡) is given by 12 x2 + x sin y = 2.
26. Solve (x2 + 2xy
through (1, 1).
y 2 )dx + (x
y)2 dy = 0 and find the solution passing
Solution. The equation is exact and the general solution is
Z x
Z y
1
2
s ds +
(x s)2 ds = x3 + x2 y xy 2 + 13 y 3 = c.
3
0
0
For x = y = 1 one has that c = 23 . Thus the solution passing through
(1, 1) is given by 13 x3 + x2 y xy 2 + 13 y 3 = 23 .
27. Solve (x2 + 2xy + 2y 2 )dx + (x2 + 4xy + 5y 2 )dy = 0. Show that there
exists a such that y = ax is a solution passing through (0, 0).
Solution. The equation is exact and the general solution is
Z x
Z y
1
5
2
s ds +
(x2 + 4xs + 5s2 )ds = x3 + x2 y + 2xy 2 + y 3 = c.
3
3
0
0
For x = y = 0 one finds c = 0. The function y = ax satisfies F (x, y) = 0
provided 1 + 3a + 6a2 + 5a3 = 0. But if we let g(a) = 1 + 3a + 6a2 + 5a3 ,
then since g(0) > 0 and g( 1) < 0, by the Intermediate Value Theorem
there exists a real number a⇤ such that g(a⇤ ) = 0. This is the unique
zero of g because g 0 (a) = 3 + 12a + 15a2 > 0.
28. Solve xdx
2y 3 dy = 0 and describe the behavior of the solutions.
Solution. Here M = x and N = 2y 3 and hence the equation is exact
(and separable). An antiderivative of ! = M dx + N dy is F (x, y) =
1 2
1 4
x
y . Thus the general solution is x2 y 4 = c. Solving for x, resp.
2
2
y, we find
(i) x = ± y 4 + c
1/2
,
resp.
(ii) y = ± x2
c
1/4
.
On the points (x, 0), x 6= 0 we have c = x2 > 0, the solutions (i) have
dx
2y 3
vertical tangents because N = 0 and solve
=
. On the points
dy
x
(0, y), y 6= 0, we have c = y 4 < 0, the solutions (ii) have horizontal
dy
x
tangents because M = 0 and solve
= 3.
dx
2y
30
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
y
x
Figure 3.3: Plot of x2
c < 0 (blue curve).
y 4 = c for c = 0 (black curve), c > 0 (red curve) and
At (0, 0) both M and N vanish. Taking c = F (0, 0) = 0 we get x = ±y 2
or, equivalently, y = ±|x|1/2 . See fig.3.3.
29. Find the number a such that (x2 + y 2 )dx + (axy + y 4 )dy = 0 is exact
and then solve it.
Solution. Since
@ 2
(x + y 2 ) = 2y,
@y
@
(axy + y 4 ) = ay
@x
then the form ! = (x2 + y 2 )dx + (axy + y 4 )dy is exact provided a = 2.
An antiderivative of (x2 + y 2 )dx + (2xy + y 4 )dy is given by
Z x
Z y
1
1
2
s ds +
(2xs + s4 )dy = x3 + xy 2 + y 5
3
5
0
0
and hence the general solution is
1 3
1
x + xy 2 + y 5 = c.
3
5
31
30. Find the coefficients ai , bi such that (x2 + a1 xy + a2 y 2 )dx + (x2 + b1 xy +
b2 y 2 )dy = 0 is exact, and solve it.
Solution. To satisfy the exactness condition, we set
@ 2
@ 2
(x + a1 xy + a2 y 2 ) =
(x + b1 xy + b2 y 2 ).
@y
@x
This amounts to solving
a1 x + 2a2 y = 2x + b1 y
i.e. (a1
2)x = (b1
2a2 )y,
8 x, y 2 R.
Then one finds a1 = 2 and b1 = 2a2 . The equation
(x2 + 2xy + a2 y 2 )dx + (x2 + 2a2 xy + b2 y 2 )dy = 0
is exact, for all real number a2 , b2 . The general solution is given by
Z x
Z y
2
s ds +
(x2 + 2a2 xs + b2 s2 )ds = 13 x3 + x2 y + a2 xy 2 + 13 b2 y 3 = c.
0
0
31. Find a function A(y) such that (2x + A(y))dx + 2xydy = 0 is exact and
solve it.
Solution. Since
@
(x + A(y)) = A0 (y),
@y
@
(2xy) = 2y
@x
the equation is exact provided A0 (y) = 2y, namely A(y) = y 2 + , 
constant.
Solving (2x + y 2 + )dx + 2xydy = 0 we find, for each constant ,
Z x
Z y
(2s + )ds +
2xsds = x2 + x + xy 2 = c.
0
0
32. Find a function B(x) such that (x + y 2 )dx + B(x)ydy = 0 is exact and
solve it.
Solution. Since
@
(x + y 2 ) = 2y,
@y
@
(B(x)y) = B 0 (x)y
@x
32
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
the equation is exact provided 2y = B 0 (x)y, namely B 0 (x) = 2 and
hence B(x) = 2x + ,  constant.
Solving (x + y 2 )dx + (2x + )ydy = 0 we find, for each constant ,
Z x
Z y
1

sds +
(2x + )sds = x2 + xy 2 + y 2 = c.
2
2
0
0
33. Show that for any di↵erentiable f (y) 6= 0 and any continuous g(y),
there exists an integrating factor µ = µ(y) for the equation f (y)dx +
(xy + g(y))dy = 0.
Solution. Here M = f (y) and N = xy + g(y). An integrating factor
µ(y) solves µ0 (y)M +µ(y)My = µ(y)Nx namely µ0 (y)f (y)+µ(y)f 0 (y) =
µ(y)y. Then
y f 0 (y)
µ0 (y) =
µ(y)
f (y)
which is separable and can be integrated yielding an integrating factor
µ(y).
34. Show that µ = x is an integrating factor for (x + y 2 )dx + xydy = 0 and
solve it.
Solution. Since
@ 2
@ 2
(x + xy 2 ) = 2xy =
(x y)
@y
@x
the equation (x2 + xy 2 )dx + x2 ydy = 0 is exact. Integrating we find
Z x
Z y
1
1
2
F (x, y) =
s ds +
x2 sds = x3 + x2 y 2 .
3
2
0
0
Thus the general solution is 13 x3 + 12 x2 y 2 = c.
35. Find an integrating factor µ(x) by multiplying the equation
(2y + x)dx + (x2
1)dy = 0
by µ(x) and determine it so that it satisfies the condition for exactness,
and then solve it.
33
Solution. Setting (2y + x)µ(x))y = (x2
2µ = µ0 (x2
Therefore,
µ0
2(1
= 2
µ
x
1)µ(x))x , we obtain
1) + 2xµ
x)
2
=
.
1
x+1
1
Solving for µ, we obtain µ = (x+1)
2 . Integrating the second term of the
exact equation
2y + x
x 1
dx +
dy = 0,
2
(x + 1)
x+1
we obtain
F (x, y) =
x 1
y + h(x).
x+1
Setting
Fx =
2y + x
(x + 1)2
h0 =
x
.
(x + 1)2
yields
Integrating by parts, we find
h(x) =
x
+ ln |x + 1|
x+1
Therefore the solutions are given by
x 1
y
x+1
36. (x + 2y)dx + (x
x
+ ln |x + 1| = c.
x+1
1)dy = 0.
M y Nx
1
1 we have
=
.
N
x 1
1
Thus an integrating factor µ(x) solves µ0 =
µ. It follows that we
x 1
can take µ = x 1.
Solution. Letting M = x + 2y, N = x
Integrating the exact equation µM dx + µN dy = 0, namely
(x + 2y)(x
1) dx + (x
1)2 dy = 0,
34
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
we find
F (x, y) =
Z
x
s(s
1)ds +
0
Z
y
1
1)2 dy = x3
3
(x
0
Thus the general solution is 13 x3
1 2
x
2
+ y(x
1 2
x + y(x
2
1)2 .
1)2 = c.
37. Solve y 2 dx + (xy + 3y 3 )dy = 0.
Solution. This equation is not exact. So, we try to see if it has an
integrating factor which is either a function of x or of y. Trying to find
µ(x), we soon find out that no such function exists. So, we try to find
a function µ(y). Here M = y 2 and N = xy + 3y 3 , we are led to solve
µ0 M + µMy = µNx
namely y 2 µ0 =
that is y 2 µ0 = (y
2y)µ
yµ.
µ
.
y
Integrating and taking the constant of integration to be 0, one has
µ0 =
ln |µ(y)| =
ln |y|
=)
µ(y) =
1
.
y
Integrating the exact form ydx + (x + 3y 2 )dy we find
Z y
F (x, y) =
(x + 3s2 )ds = xy + y 3 = c.
0
38. Solve (1 + y 2 )dx + xydy = 0.
Solution. Here M = 1 + y 2 and N = xy, and hence an integrating
factor µ(y) can be found by solving
µ0 M + µMy = µNx
that is µ0 =
y
µ.
1 + y2
Integrating, one solution is given by
1
µ(y) = p
.
1 + y2
Now the form
!=
p
xy
1 + y 2 dx + p
dy
1 + y2
35
is exact. Integrating,
Z x
Z y
p
xs
p
F (x, y) =
ds+
dy = x+x 1 + y 2
1 + s2
0
0
or
x=x
p
1 + y 2 = c,
c
x= p
,
1 + y2
which is the general solution.
39. Solve x0 = (x + 2t)/t, t 6= 0.
Solution. The equation is homogeneous. Setting x = tz we find
z + tz 0 =
tz + 2t
=z+2
t
=)
tz 0 = 2
=)
z0 =
2
t
Hence z = c + ln t2 and x = tz = t(c + ln t2 ), t 6= 0.
40. Solve x0 = tx/(t2 + x2 ).
Solution. The equation is homogeneous. Setting x = tz we find
z + tz 0 =
t2 z
t2 + t2 z 2
Then
tz 0 =
namely z + tz 0 =
z
1 + z2
z=
z
1 + z2
z3
1 + z2
and hence either z = 0 or
1 + z2 0
·z =
z3
1
,
t
t 6= 0.
Integrating
Z
1 + z2
· dz =
z3
Z
dt
1
= ln
+ c,
t
t
t 6= 0.
The integral on the left hand side can be split into
Z
Z
Z
1 + z2
dz
dz
1
· dz =
+
=
+ ln |z| + c
3
3
z
z
z
2z 2
36
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
and hence
1
1
+ ln |z| = ln
+ c,
2
2z
t
t 6= 0.
Finally form x = tz, that is z = x/t (t 6= 0), we infer that either x = 0
or
t2
x
1
+ ln
= ln
+ c, t 6= 0.
2
2x
t
t
41. Solve x0 =
3x2
2t2
tx
.
Sol. Since the equation is homogeneous, we let x = tz and obtain
z + tz 0 =
3t2 z 2 2t2
3z 2 2
=
t2 z
z
whereby
3z 2 2
2z 2 2
z=
z
z
Separating the variables and integrating, we obtain
tz 0 =
4 ln |t|
Replacing z by
ln |2z 2
2| = k,
namely
ln
t4
2z 2
x
and simplifying, we obtain
t
t6
2x2
2t2
= c.
x2 + t 2
.
2tx
Sol. We let x = tz and obtain
42. Solve x0 =
z + tz 0 =
whereby
tz 0 =
t2 z 2 + t2
z2 + 1
=
2t2 z
2z
z2 + 1
2z
z=
1
z2
2z
2
= k.
37
Separating the variables and integrating, we find
z 2 | = k.
ln |t| + ln |1
x
and simplifying, we obtain
t
✓
◆
x2
t 1
=c
namely
t
t2
Replacing z by
43. Solve x0 =
x2
= c.
t
x t+1
.
x t+2
Solution. Substituting x t = v in the di↵erential equation, we obtain
v 0 + 1 = v+1
or
v+2
v+1
1
v0 =
1=
.
v+2
v+2
Integrating this separable equation we find 12 v 2 + 2v = t + c. Substituting (x t) for v, we obtain
1
(x
2
44. Solve x0 =
t)2 + 2(x
t) + t = c,
i.e.
t)2 + 2x
t = c.
x t
.
x t+1
Solution. Letting x
v0 =
v
v+1
t = v we obtain v 0 + 1 =
1
1
v + ln |2v + 1| =
2
4
2(x
2v + 1
v+1
1=
1
Therefore 12 [1 + 2v+1
]dv =
simplifying, we obtain
i.e.
v
v+1
or
v+1
dv =
2v + 1
dt
dt. Integrating, replacing v by x
2v + ln |2v + 1| =
t + c,
t) + ln |2(x
t) + 1| =
2(x + t) + ln |2(x
45. Solve x0 =
1
(x
2
4t + c
t) + 1| = c
x+t+1
.
x t+1
Solution. Substituting x = v + k and t = u + h, we have
dv
dx
=
=
du
dt
v+k+u+h+1
.
v+k u h+1
4t + c
t and
38
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
Solving the system
we find h = 0, k =
(
k h+1 =0
k+h+1 =0
1, yielding v = x + 1, u = t and
dv
=
du
v+u
.
v u
Now, substituting v = uw in this homogeneous equation, simplifying
and separating the variables, we obtain
du
w 1
+ 2
dw = 0.
u
w +1
Integrating, we obtain
1
ln[w2 + 1] arctan w = c.
2
v
x+1
Now we substitute t for u and =
for w obtaining
u
t


1
(x + 1)2
x+1
ln |t| + ln
+
1
arctan
= c.
2
t2
t
ln |u| +
46. Solve the Cauchy problem x0
solutions relative to a.
x = tx2 , x(0) = a > 0 and describe the
Solution. This is a Bernoulli equation discussed in Example 3.5.1 of
the text. By uniqueness, the solution is positive. We have found that
the general solution is given by
x(t) =
ce
t
1
+1
t
,
ce
t
+1
t 6= 0.
Inserting the initial condition x(0) = a > 0 we find
1
,
c 6= 1,
c+1
1
1 a
which yields c =
1=
. Thus the solution is
a
a
1
a
xa (t) =
=
t
1 a t
(1 a)e + a(1
e +1 t
a
a=
t)
.
39
Since the solution has to be positive, we have to require (1
a(1 t) > 0.
a)e
t
+
The domain of definition of xa depends on a. First of all, since the
solution has to be positive, we have to require
(1
a)e
t
+ a(1
t) > 0.
To solve this inequality we have to distinguish if 0 < a < 1, a = 1 or
a > 1 (the case a = 1 has been discussed in the Example 3.5.1). In the
former case, solving (1 a)e t > a(t 1)
x
x
(a) (1
a)e
t
> a(t
α
t
α
1), 0 < a < 1
β
t
(b) (1
a)e
t
> a(t
1), a > 1
Figure 3.4
we find t < ↵, where ↵ > 1 is such that (1 ↵)e ↵ = ↵(t 1), see fig.
3.4a. If a > 1 then the equation (1 a)e ↵ = a(t 1) has two solutions
↵ < 0 < and we find ↵ < t < , see fig. 3.4b.
47. Find the nontrivial solutions of x0 + 2tx =
4tx3 .
Solution. This is a Bernoulli equation with k = 2, p = 2t and q =
1
Setting z = 2 , x 6= 0, z solves
x
z 0 4tz = 8t.
This is a linear equation whose general solution is
z = ce2t
and hence
2
2
1
1
x = ±p = ±p 2
z
ce2t
2
.
4t.
40
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
48. Find the nontrivial solutions of x0
tx = x2 .
Solution. This is a Bernoulli equation with k = 1, p = t and q = 1.
Setting z = x 1 we find z 0 = x 2 x0 = x 2 (tx + x2 ) = tx 1 1 =
tz 1. To find the general solution of the linear equation z 0 +tz = 1,
2
we multiply both sides by the integrating factor et /2 yielding
2 /2
2
2
(z 0 + tz) = et /2 =) (et /2 z)0 =
R t2 /2
2
Integrating, et /2 z =
e
+ c and therefore
✓ Z
◆
t2 /2
t2 /2
z=e
e dt + c
et
et
2 /2
which gives the solution by setting x = z 1 .
49. Show that the circle x2 + t2 = 1 is the singular solution of x02 =
x2 + t2 1.
Solution. Set F (x, x0 ) = x02 x2 t2 + 1. According to the definition,
x = (t) is a singular solution of F (x, x0 ) = 0 if it solves the system
F = 0, Fx0 = 0, namely
⇢ 02
x = x2 + t2 1,
2x0 = 0,
and is such that Fx ( (t), 0 (t)) 6= 0. From the second equation of the
2
2
system we find x0 = 0 and
p hence the first equation yields 0x +t 1 = 0,
2
whereby x = (t) = ± 1 t . We also have Fx ( (t), (t)) = 2 (t),
which di↵erent from zero for t 6= ±1. In this sense we can say that
x2 + t2 = 1 defines the singular solution.
50. Solve x02 = 4(1
x) a show that x = 1 is the singular solution.
p
Solution. This is equivalent to the pair of equations x0 = ±2p 1 x
which are separable autonomous. Either x = 1 or, dividing by 2 1 x,
we find
dx
p
= ±dt
2 1 x
Integrating, one has
Z
p
dx
p
= ±t + c =)
1 x = ±t + c
2 1 x
41
Squaring both sides, we get 1 x = (c ± t)2 , that is x = 1 (c ± t)2 ,
a family of parabolas with vertex on the line x = 1. Thus x = 1 is the
envelope of the parabolas. Setting F (x, x0 ) = x02 4(1 x) and solving
the system F = 0, Fx0 = 0, namely
⇢ 02
x = 4(1 x)
2x0 = 0,
we find x = 1. Notice that Fx = 4 6= 0. Then x = 1 is the singular
solution.
51. Find a singular solution of x02
tx0 + x = 0.
Solution. One can solve the system F = 0, Fx0 = 0. Or else, one can
check that x = h(t, c) = ct c2 solves F = 0 for all constants c. Solving
@c h(t, c) = 0 we find t = 2c. Substituting c = 2t into x = h(t, c) we
have x = 14 t2 , the envelope of the family x = ct c2 . The function
x = 14 t2 is a solution and hence it is a singular solution.
52. Solve the Clairaut equation x = tx0
x02 and find the singular solution.
Solution. Putting x0 = c we find x = ct c2 . The inverse of g 0 (p) =
2p is h(s) = s/2. Hence the singular solution is
✓ ◆2
t
t2
= .
2
4
t2
(t) = th( t) + g(h( t)) =
2
0
53. Solve the Clairaut equation x = tx0 + ex and find the singular solution.
Solution. The general solution is x = ct+ec . The inverse of g 0 (p) = ep
is h(s) = ln s, with s > 0. Then the singular solution is
(t) = th( t) + g(h( t)) = t ln( t) + eln(
t)
= t ln( t)
t,
t < 0.
54. Solve the Clairaut equation x = tx0 ln x0 and find the singular solution.
Solution. Here the function g(p) = ln p is defined on R+ = {p 2 R :
p > 0} and hence
x(t) = ct ln c,
c > 0.
42
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
x
t
Figure 3.5: Solutions of x = tx0
x = t ln( t) t, t < 0
0
ex . The red curve is the singular solution
Let us find the singular solution. Here g 0 (p) = 1/p, p > 0 and hence
h(t) = 1/t, t > 0. Then the singular solution is given by
(t) = th( t) + g(h( t)) = 1
ln
1
= 1 + ln t,
t
t > 0.
x
O
t
Figure 3.6: Solutions of x = tx0
x = 1 + ln t
ln x0 . The red curve is the singular solution
55. Solve the Clairaut equation x = tx0 + x10 and find the singular solution.
Solution. Since g(p) = 1/p then x = ct+ 1c , c 6= 0. To find the singular
solution, let us find directly the envelope of the family of straight lines
43
x = ct + 1c . Taking the derivative with respect to c we get t c 2 = 0.
p
Then c = ±t 1/2 with t 0, whereby x = ±t · t 1/2 + ±t 11/2 = ±2 t,
or 4t = x2 .
56. Find ↵, such that x(t) = ↵t + solves the D’Alambert-Lagrange
equation x = th(x0 ) + g(x0 ), h, g 2 C(I).
(a) Show that the equation x = t(1 + x0 ) + x0 has no solution of the
form x = ↵t + .
(b) Solve the equation by setting x0 = z.
Solution. Substituting x = ↵t + and x0 = ↵ into the equation, we
find ↵t + = th(↵) + g(↵). Since this is to hold for all t 2 I we infer
↵ = h(↵) and = g(↵). Thus, for any ↵ such that ↵ = h(↵), the
solutions are
x(t) = h(↵)t + g(↵).
Notice that if h(x0 ) = x0 we find the Clairaut equation and we can take
any constant ↵ yielding x = ↵t + g(↵).
(a) Substituting x = ↵t+ into the equation we find ↵t+ = t(1+↵)+↵
whereby ↵ = 1 + ↵, which is impossible.
(b) Setting x0 = z we find x = t(1 + z) + z. Di↵erentiating, we find
x0 = z = 1 + z + tz 0 + z 0 , that is (t + 1)z 0 = 1. Integrating we obtain
z(t) =
ln |t + 1| + c,
t 6=
1,
c 2 R,
whereby
x = t + (1 + t)z = t + (1 + t) (c
ln |t + 1|) ,
t 6=
1,
c 2 R.
44
CHAPTER 3. SOLUTIONS TO EXERCISES OF CHAPTER 3
Chapter 4
Solutions to exercises of
Chapter 4
1. Show that the Cauchy problem x00 = x|x|, x(0) = a, x0 (0) = b, has a
unique solution, for all a, b 2 R.
Solution. The function f (x) = x|x| is such that f 0 (x) = 2|x|, which is
continuous. Using Lemma 2.4.2 it follows that f is locally lipschitzian.
Then we can apply the Existence and Uniqueness Theorems 4.2.1 and
4.2.2.
2. Show that the Cauchy problem x00 = max{0, x|x|}, x(0) = a, x0 (0) = b,
has a unique solution, for all a, b 2 R.
Solution. The function f (x) = max{0, x|x|} is such that f 0 (x) =
max{0, 2x}, which is continuous.
3. Show that for p 2 the Cauchy problem x00 = |x|p , x(0) = a, x0 (0) = b,
has a unique solution, for all a, b 2 R.
Solution. If p
2 the function f (x) = |x|p is di↵erentiable and
0
f (x) =
⇢
pxp 1 , if x 0
pxp 1 , if x < 0
which is is continuous.
45
46
CHAPTER 4. SOLUTIONS TO EXERCISES OF CHAPTER 4
4. Let x, y be the unique solution
8 0
< x
y0
:
x(0)
of
=y
= x
= 0, y(0) = 1
Show that x, y verify x2 + y 2 ⌘ 1.
Solution. Di↵erentiating and using the equations x0 = y, y 0 = x we
find
d 2
(x + y 2 ) = 2xx0 + 2yy 0 = 2xy 2yx = 0
dt
2
Hence x (t) + y 2 (t) ⌘ c, c a constant. For t = 0 we get c = x2 (0) +
y 2 (0) = 0 + 1 = 1.
5. Do the same as in Problem 4 when x, y solve
8 0
= ✓y
< x
0
y
= ✓x
:
x(0) = 0, y(0) = 1
Solution. Di↵erentiating and using the equations x0 = ✓y, y 0 =
we find
d 2
(x + y 2 ) = 2xx0 + 2yy 0 = 2✓xy 2✓yx = 0
dt
and the conclusion follows in the same way.
6. Prove that the solutions x, y of
⇢ 0
x = Hy (x, y)
y 0 = Hx (x, y)
where H : R ⇥ R 7! R is smooth, satisfy H(x(t), y(t)) = c.
Solution. One has
dH(x, y)
= Hx (x, y)x0 + Hy (x, y)y 0
dt
= Hx (x, y)Hy (x, y) Hy (x, y)Hx (x, y) = 0.
7. Find the 2nd order equation a solution of which is x(t) = et + e t .
Solution. One has x00 = et + e
t
= x.
✓x
47
8. Same for x = tet .
Solution. x0 = et + tet and x00 = 2et + tet =
x
00
✓
2x
+ x. Hence
t
◆
2
+ 1 x = 0.
t
9. Let x be the solution of x00 + x = 0, x(0) = 0, x0 (0) = 1. Prove that
x = sin t.
Solution. The function x(t) = sin t verifies x00 = (sin t)00 = sin t = x.
Moreover, x(0) = 0 and x0 (0) = cos 0 = 1. By uniqueness, x(t) = sin t.
10. Let x be the solution of x00 + 4x = 0, x(0) = 1, x0 (0) = 0. Prove that
x = cos 2t.
Solution. As before, x(t) = cos 2t verifies x00 = 4x. Moreover,
x(0) = 1 and x0 (0) = 2 sin 0 = 0. By uniqueness, x(t) = cos 2t.
11. Let f (t, x) be T -periodic with respect to t and let x(t) be a solution
of x00 = f (t, x), defined for all t 2 R, such that x(0) = x(T ) and
x0 (0) = x0 (T ). Show that x(t) is T -periodic.
Solution. Setting z(t) = x(t + T ), one has z 00 (t) = x00 (t + T ) = f (t +
T, x(t + T )) = f (t + T, z(t)) = f (t, z(t)). Moreover, z(0) = x(T ) = x(0)
and z 0 (0) = x0 (T ) = x0 (0). By uniqueness it follows that z(t) = x(t)
whereby x(t + T ) = x(t).
12. Let f : R 7! R be smooth and and let x : R 7! R satisfy x0000 = f (x),
x0 (0) = 0, x000 (0) = 0. Show that x(t) is even.
Solution. If z(t) = x( t) one finds z 0000 (t) = x0000 ( t) = f (x( t)) =
f (z(t)). Furthermore, z(0) = x(0), z 0 (0) = x0 (0) = 0, z 00 (0) =
x00 (0), z 000 (0) = x000 (0) = 0. Thus, by uniqueness, z(t) = x(t) namely
x( t) = x(t).
13. Show that if f is continuous and f (x) > 0 for all x 2 R, then the
solutions of x000 = f (x), have an inflection point at t = 0.
Solution. One has x000 (t) = f (x(t)) > 0 and hence x00 (t) is strictly
increasing. Thus x00 (t) > x00 (0) = 0 for t > 0 and x00 (t) < x00 (0) = 0 for
t < 0.
48
CHAPTER 4. SOLUTIONS TO EXERCISES OF CHAPTER 4
Chapter 5
Solutions to exercises of
Chapter 5
A1. Show that x1 = t3
t2 and x2 = t3
3t are linearly independent.
Solution. Suppose that c1 (t3 t2 ) + c2 (t3 t) = 0 for all t. Then we would
have (c1 + c2 )t3 c1 t2 c2 t = 0 for all t. Therefore we must have all the
coefficients equal to zero, i.e. c1 + c2 = c1 = c2 = 0.
A2. Consider the functions f (t) = sin t and g(t) = t2 .
a) Using the definition of linear independence, explain why they are linearly independent.
b) Using a Wronskian argument, explain why they are linearly independent.
c) Explain why they cannot be solutions of a di↵erential equation x00 +
p(t)x0 + q(t)x = 0, where p and q are continuous functions.
Solution. a) Suppose there exist constants c1 and c2 such that c1 sin t +
c2 t2 = 0 for all t. Then substituting t = ⇡, we obtain c2 = 0, which implies
that c1 sin t = 0 and hence c1 = 0, by letting, for example, t = ⇡2 .
b) Di↵erentiating the identity c1 sin t + c2 t2 = 0 we find c1 cos t + 2c2 t = 0,
for all t. The determinant of the system
(
c1 sin t + c2 t2 = 0,
c1 cos t + 2c2 t = 0,
49
50
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
is the Wronskian W (sin t, t2 ) which is equal to 2t sin t t2 cos t. The equations
of the systems hold for all t. For t = ⇡2 one has W ( ⇡2 ) = ⇡ 6= 0. Then
c1 = c2 = 0 and the functions are linearly independent.
c) Since W (0) = 0 and W ( ⇡2 ) = ⇡ 6= 0, they cannot be solutions since
Abel’s Theorem implies that the Wronskian of two solutions is either always
0 or never 0.
A3. Show that if x1 and x2 are linearly independent, then so are their linear
linear combinations z1 = 2x1 + 3x2 and z2 = 2x1 3x2 .
Solution. Suppose that c1 (2x1 (t) + 3x2 (t)) + c2 (2x1 (t) 3x2 (t)) = 0 for all t
in some interval I. Writing the previous identity as (2c1 + 2c2 )x1 (t) + (3c1
3c2 )x2 (t) = 0, we see that c1 and c2 must be solutions to the system
⇢
2c1 + 2c2 = 0,
3c1 3c2 = 0,
which has only the trivial solution c1 = 0 = c2 since the determinant of the
coefficients is 2 · ( 3) 2 · 3 = 12 6= 0.
A4. (a) Prove that if the Wronskian of two di↵erentiable functions f (t) and
g(t), not necessarily solutions of di↵erential equations, is nonzero at one point
of an interval I, then they are linearly independent.
(b) Prove that if they are linearly dependent, then their Wronskian is identically equal to 0.
Solution. (a) Suppose that W (t0 ) 6= 0. Then the algebraic system
⇢
c1 f (t0 ) + c2 g(t0 ) = 0
c1 f 0 (t0 ) + c2 g 0 (t0 ) = 0
has a unique solution in c1 and c2 , and must therefore be c1 = 0 = c2 .
(b) If the assertion were false, then their Wronskian would be nonzero
at some point. But that would imply, by part (a), that they are linearly
independent; contradicting the assumption.
A5. Show that x1 = tan t and x2 = sin t are linearly independent on the
interval (0, ⇡).
51
Solution. Since
the Wronskian W (t) = sin t sin t sec2 t, we see that
p
2
W ( ⇡4 ) =
6= 0. Therefore they are linearly independent (see Problem
2
A4).
A6. Solve the initial value problem
W (t2 + 1, f (t)) = 1, f (0) = 1.
Solution. Replacing W (t2 + 1, f (t)) by (t2 + 1)f 0 (t)
equivalent problem
(t2 + 1)f 0 (t)
2tf (t), we obtain the
2tf (t) = 1, f (0) = 1.
Using the method of integrating factor, we solve this first order linear equation, obtaining
✓
◆0
f (t)
1
= 2
2
t +1
t +1
and hence
✓
f (t)
2
t +1
◆
= arctan t + c.
Substituting the initial value t = 0, we obtain c = 1, which gives the solution
f (t) = (t2 + 1)(arctan t + 1).
A7 Show that if x1 and x2 be linearly independent functions and z(t) is a
function such that z(t) 6⌘ 0 in I, then zx1 and zx2 are linearly independent
on I.
Solution. One has W (zx1 , zx2 ) = zx1 (z 0 x2 + zx02 ) zx2 (z 0 x1 + zx01 ) =
z 2 x1 x02 z 2 x01 x2 = z 2 W (x1 , x2 ). If zx1 and zx2 are linearly dependent,
W (zx1 , zx2 ) ⌘ 0. By hypothesis, there exists a number t̄ in I such that
z(t̄) 6= 0. Hence W (zx1 , zx2 ) ⌘ 0 implies W (x1 , x2 ) ⌘ 0 and thus x1 , x2
would be linearly dependent (see Problem A4), a contradiction.
A8. Give an example to show that the following statement is false: if two
functions f1 and f2 are linearly independent in an interval I, then they are
also independent in any subinterval J of I.
52
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
Solution. One way could be to define x1 (t) and x2 (t) as follows: x1 (t) = t2
for all t in the interval [ 2, 2] and define x2 (t) as:
⇢ 2
t
if t 2 [0, 2]
x2 (t) =
2
t if t 2 [ 2, 0].
Then, they are certainly linearly dependent in the interval J = [0, 1] since
x2 = x1 on this interval. However, on the larger interval [ 2, 2] they are
linearly independent. To see this, suppose there exist nonzero constants c1
and c2 such that c1 x1 (t) + c2 x2 (t) = 0 in I. Then substituting the numbers
t = 1 and t = 1, we obtain the algebraic equations
⇢
c1 + c2 = 0
c1 c2 = 0
which can be true only if c1 = 0 = c2 .
A9. Show that if x1 and x2 are linearly dependent on an interval I, then
they are linearly dependent in any subinterval J of I.
Solution. This is obvious since c1 x1 (t) + c2 x2 (t) = 0 for all t in I would
imply that c1 x1 (t) + c2 x2 (t) = 0 for all t in any subinterval of I.
A10. Show that if two solutions of a second order homogeneous di↵erential
equation with continuous coefficients on I have a common zero then all their
zeros are in common.
Solution. Let x1 and x2 be solutions such that x1 (t0 ) = 0 = x2 (t0 ) for some
number t0 in the domain. Then, W (x1 , x2 )(t0 ) = x1 (t0 )x02 (t0 ) x01 (t0 )x2 (t0 ) =
0. Therefore, x1 and x2 are linearly dependent, which means that one of them
is a constant multiple of the other, i.e. x1 = kx2 . Therefore, if one of them
vanishes at a point, then so does the other.
0
A11. Let x1 and x2 be two solutions of x00 + xt + q(t)x = 0, t > 0, where q(t)
is a continuous function. Given that W (6) = 7, find W (7).
R
1
Solution. By Abel’s Theorem, W (t) = ce t dt = ce ln t = c/t, where c is
a constant. Since W (x1 , x2 )(6) = 7, we must have 7 = c/6, thus obtaining
c = 42. This means that W (t) = 42/t. Therefore, W (7) = 42
= 6.
7
53
B1. 2x00 + x0
x = 0.
Solution: The corresponding characteristic equation 2m2 + m 1 = 0 has
1
roots m = 1, 12 . Therefore the general solution is x = c1 e t + c2 e 2 t .
B2. x00 + 2x0 + 2x = 0.
Solution. The roots of the characteristic equation are the complex numbers
1 ± i, which gives the general solution x = c1 e t sin t + c2 e t cos t.
B3. x00 + 8x0 + 16x = 0.
Solution. Since the discriminant is 0, the corresponding characteristic equation m2 +8m+16 = 0 has one double (i.e. repeated) root m = 4. Therefore
the general solution is x = c1 e 4t + c2 te 4t .
B4. x00 + 2x0
15x = 0, x(0) = 1, x0 (0) = 1.
Solution. The corresponding characteristic equation is m2 + 2m 15 = 0,
which has roots m = 3, 5. Thus the general solution is x = c1 e3t + c2 e 5t .
Substituting the initial values gives rise to the algebraic system
⇢
c1 + c2
= 1,
3c1 5c2 = 1,
which has the solution c1 = 34 and c2 = 14 . Consequently, the solution to the
problem is x = 34 e3t + 14 e 5t .
B5. x00
3x00 + 2x = 0, x(1) = 0, x0 (1) = 1 .
Solution. The corresponding characteristic equation is m2 3m + 2 = 0,
which has roots m = 1, 2. Thus the general solution is x(t) = c1 et + c2 e2t .
Now, imposing the initial conditions we find the system
⇢
ec1 + e2 c2 = 0
ec1 + 2e2 c2 = 1
Subtracting the second equation from the first, one has e2 c2 = 1, that is
c2 = e 2 . Substituting in the first equation we find e c1 + 1 = 0, that is
54
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
c1 = e 1 . So the solution of the initial value problem is x(t) =
e 2 e2t = et 1 + e2t 2 .
B6. 4x0 + 2x00 =
e 1 et +
5x, x(0) = 1, x0 (0) = 1 .
Solution. The characteristic
equation corresponding to 2x00 + 4x0 + 5x = 0
p
has solutions m = 1 ± 26 i. Therefore, the general solution is
p
p
6
6
x = c1 e t sin
t + c2 e t cos
t.
2
2
Substituting the initial values and solving the algebraic system for c1 and c2 ,
4
we obtain c1 = p and c2 = 1. Thus the desired solution is
6
p
p
4
6
6
x = p e t sin
t + e t cos
t.
2
2
6
B7. x00
6x0 + 9x = 0, x(0) = 0, x0 (0) = 1.
Solution. Since m = 3 is a double root of he characteristic equation
m2 6m + 9 = 0, the general solution is x = c1 e3t + c2 te3t . Substituting
the initial values gives us c1 = 0 and c2 = 1. The required solution is therefore x = te3t .
B8. Show that for
0, x00 + x0
x = 0 will always have some solutions
that do not approach 0 as t ! +1.
p
1± 1+4
Solution. The roots of the characteristic equation are m =
.
2
We take 2 cases.
Case 1. If = 0, then the general solution is x = c1 e t + c2 . So, the
assertion follows if c1 = 0 and c2 6= 0.
p
Case 2. If > 0, then one root is m = 1+ 21+4 > 0 and the assertion
follows.
B9. For what values of
will all solutions of
x00 + x0
x=0
55
go to 0 as t goes to 1?
Solution. The assertion is false for
0 (see the preceding problem). For
any < 0, all solutions tend to 0 as t ! 1. To show this, we consider 3
cases involving
p the roots of of the characteristic equation which are given by
1± 1+4
m1,2 =
.
2
1
1
Case 1. 1+4 = 0. In this case, the general solution is x = c1 e 2 t +c2 te 2 t
which implies that all solutions tend to 0 as t ! +1.
p
Case 2. 1 + 4 > 0. Then, since < 0, it follows that m1,2 = 1± 21+4 <
0 and the assertion follows.
p
Case 3. 1 + 4 < 0. In this case, if we let 1+4
= i✓, then the general
2
solution is
1
x = e 2 t (c1 sin ✓t + c2 cos ✓t)
and again the assertion follows.
B10. Show that all the solutions of x00 + 4x0 + kx = 0 go to 0 as t ! +1
if and only if k > 0.
p
Solution. The roots of the characteristic equation are m1,2 = 2 ± 4 k.
For k > 0, there are 3 cases:
Case 1. If k > 4 the roots are imaginary, in which case the general
solution will look like c1 e 2t sin t + c2 e 2t cos t.
Case 2. If k = 4, we have a double root m = 2, in which case the
general solutions is c1 e 2t + c2 te 2t .
p
Case 3. If 0 < k < 4, the roots are m1,2 = 2 ± 4 k < 0, in which
case the general solutions is c1 em1 t + c2 tem2 t .
In all three cases the assertion follows. p
On the other hand, if k < 0, then 2 + 4 k > 0, which implies that
the corresponding solution tends to 1 as t ! 1. If k = 0, all the solutions
x = c 6= 0 do not go to zero.
B11. Show that the equation x00 + bx0 + cx = 0, has infinitely many solutions
and none of them, except the trivial solution, can have a maximum or a
minimum point on the t-axis.
Solution. If a solution x has a maximum or a minimum on the t-axis, then
56
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
x and x0 would be zero at the same point, which would mean that x is the
trivial solution, by uniqueness.
B12. Find a second order linear homogeneous equation whose corresponding
characteristic equation has m = 3 7i as one of its roots.
Solution. We recall that complex roots come in pairs, i.e. if ↵ + i is a root
of the polynomial, then so is its conjugate ↵
i. Therefore, the characteristic equation is [m (3 7i)][m (3 + 7i)] = m2 6m + 58. the corresponding
di↵erential equation is x00 6x0 + 58x = 0.
B13. Show that any solution of x00 + 5x0 + 6x = 0 tends to zero as t ! +1.
Solution. The solutions of the characteristic equation m2 + 5m + 6 = 0
are m1 = 2, m2 = 3. Then the general solution is x(t) = c1 e 2t + c2 e 3t ,
which tends to zero as t ! +1 for every c1 , c2 .
B14. Show that if p > 0 then any solution of x00 + px0 = 0 tends to a constant
as t ! +1, while if p < 0 only constant solutions tend to constants.
Solution. The solutions of the characteristic equation m2 + pm = 0 are
m1 = 0, m2 = p and the general solution is x(t) = c1 + c2 e pt . If p > 0 one
finds
lim c1 + c2 e pt = c1 .
t!+1
On the other hand, if p < 0 one has that the
lim
t!+1
c1 + c2 e
pt
is constant if and only if c2 = 0.
B15. Find a such that the solution of x00 + x0
tends to zero as t ! +1.
2x = 0, x(0) = a, x0 (0) = 1
Solution. The solutions of the characteristic equation m2 + m 2 = 0 are
m1 = 1, m2 = 2 and hence the general solution is x(t) = c1 et + c2 e 2t .
Clearly x(t) ! 0 as t ! +1 if and only if c1 = 0 so that we have
x(t) = c2 e 2t . Then x(0) = 0 yields c2 = a, while x0 (0) = 1 implies
57
x0 (0) = 2e0 = 2c2 = 1. It follows that 2a = 1, that is a =
that one has that c2 = 12 and thus the solution becomes x(t) =
B16. Show that all solutions of x00
and unbounded on [0, 1).
1
. Notice
2
1
e 2t .
2
2x0 + 2x = 0 are bounded on ( 1, 0],
Solution. The solutions of the characteristic equation m2 2m + 2 = 0 are
m1,2 = 1 ± i and hence the general solution is x(t) = et (c1 sin t + c2 cos t).
Then for all c1 , c2 one has
lim et (c1 sin t + c2 cos t) = 0,
t! 1
lim et (c1 sin t + c2 cos t) = +1,
t!+1
and hence x(t) is bounded on ( 1, 0], while on [0, 1) the solution et (c1 sin t+
c2 cos t) is unbounded.
B17. Find a, b such that the solutions of x00
functions.
2ax0 + bx = 0 are oscillating
p
Solution. The roots of the characteristic equation are m1,2 = a ± a2 b.
If a2 b > 0, these roots are real, distinct and the general solution is x(t) =
c1 em1 t + c2 em2 t . If a2 b = 0 we have a double root m = a and the general
solution is x(t) = c1 eat + c2 teat In both cases, the equation x(t) vanishes has
at most once. Thus the solutions have at most one zero and hence do not
oscillate.
p
If a2 b < 0, the roots are a ± i✓, where ✓ = b a2 , and the general
solution is x(t) = eat (c1 sin ✓t + c2 cos ✓t) which oscillates because it has infinitely many zeros.
B18. Find
6= 0 such that the boundary value problem x00 +
x(0) = x(⇡) = 0, has nontrivial solutions.
2
x = 0,
Solution. The general solution is x(t) = c1 sin t + c2 cos t. The condition
x(0) = 0 yields c2 = 0 and hence x(t) = c1 sin t. Then x(⇡) = 0 implies
0 = c1 sin ⇡. Excluding c1 = 0 that gives rise to the trivial solution, it
follows sin ⇡ = 0 and thus ⇡ = k⇡, that is = k, k = 1, 2, ... Notice that,
for each k, the problem has infinitely many solutions given by c1 sin kt.
58
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
B19. Find a 6= b such that the boundary value problem x00 + x = 0,
x(a) = x(b) = 0, has nontrivial solutions.
Solution. The general solution is x(t) = c1 sin t + c2 cos t. Imposing the
conditions x(a) = x(b) = 0 we find
⇢
c1 sin a + c2 cos a = 0
c1 sin b + c2 cos b = 0
The system has the trivial solution c1 = c2 = 0. According to the Kramer
rule, the system has a nontrivial solution if and only if the determinant of
the system is zero, that is
sin a cos a
sin b cos b
= sin a cos b
cos a sin b = sin(a
b) = 0
Then for any a, b such that a b = k⇡, k = 1, 2, ..., the problem has nontrivial
solutions.
B20. Show that the boundary value problem x00
has only the trivial solution.
x = 0, x(0) = x(1) = 0,
Solution. The general solution is x(t) = c1 et + c2 e t . The conditions
x(0) = x(1) = 0 yield
⇢
c1 + c2
=0
c1 e + c2 e 1 = 0
The determinant of the system is
1 1
e e 1
=e
1
e.
Since e 1 e 6= 0, the Kramer rule implies that the solution of the system is
c1 = c2 = 0 which yields the trivial solution x(t) ⌘ 0.
B21. Sow that the problem x00 + x0
only the trivial solution x(t) = 0.
2x = 0, x(0) = 0, limt!+1 x(t) = 0 has
Solution. The general solution is x(t) = c1 et + c2 e 2t . The condition
x(0) = 0 yields c1 = c2 and hence we find x(t) = c1 et c1 e 2t which
59
has infinite limit unless c1 = 0.
B22. Solve x00
2x0 + 5x = 0, x(0) = 1, x(⇡/4) = 0.
Solution. The general solution is x(t) = et (c1 sin 2t + c2 cos 2t). The conditions x(0) = 1 yields c2 = 1, while x(⇡/4) = 0 gives e⇡/4 c1 = 0, that is
c1 = 0. Thus the solution is
x(t) = et cos 2t.
B23. Find ✓ such that x00
trivial solution.
2x0 + 5x = 0, x(0) = 0, x0 (✓) = 0, has only the
Solution. The general solution is x(t) = et (c1 sin 2t + c2 cos 2t) and x(0) = 0
implies c2 = 0 and hence x(t) = c1 et sin 2t. Then x0 (✓) = c1 e✓ sin 2✓ +
2c1 e✓ cos 2✓. Solving c1 e✓ sin 2✓ + 2c1 e✓ cos 2✓ = 0 we get
c1 e✓ (sin 2✓ + 2 cos 2✓) = 0
If ✓ is such that sin 2✓ + 2 cos 2✓ 6= 0 then c1 = 0 and the only solution is
x(t) ⌘ 0. Otherwise any x(t) = c1 et sin 2t solves the problem.
B24. Solve x00 + 2x0 = 0, x(0) = 0, limt!+1 x(t) = a.
Solution. The solutions of x00 + 4x0 = 0, x(0) = 0, are x(t) = c
Then
lim x(t) = c.
ce
2t
.
t!+1
Thus c = a and the solution is x(t) = a
C1. Solve x00
ae
2t
.
4x = t.
Solution. The general solution is of the associated homogeneous equation
x00 4x = 0 is x(t) = c1 e2t + c2 e 2t . Let us find a particular solution h(t) in
the form h(t) = at. From h00 4h = t we deduce 4at = t that is a = 14 .
Thus the general solution is
x(t) = c1 e2t + c2 e
2t
1
t.
4
60
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
C2. Solve x00
4x = 4t2 .
Solution. We seek a particular solution h(t) in the form h(t) = At2 +Bt+C.
The equation h00 4h = 4t2 yields h00 4h = 2A 4At2 4Bt 4C = 4t2
whereby A = 1, B = 0, C = 12 . Thus x = c1 e2t + c2 e 2t t2 12 .
C3. Solve x00 + x = t2
2t.
Solution. We seek a particular solution g(t) in the form g(t) = At2 +
Bt + C. Then g 00 + g = 2A + At2 + Bt + C = At2 + Bt + C + 2A and
g 00 + g = t2 2t yields A = 1, B = 2, C = 2A = 2. It follows that
g = t2 2t 2 and the solution is
x(t) = c1 sin t + c2 cos t + t2
2t
2.
C4. Solve x00 + x = 3t2 + t.
Solution. Setting g(t) = At2 + Bt + C, from g 00 + g = 3t2 + t it follows
At + Bt + C + 2A = 3t2 + t whence A = 3, B = 1, C = 2A = 6. Then
the solution is
x(t) = c1 sin t + c2 cos t + 3t2 + t 6.
2
C5. Solve x00
g
00
x=e
3t
.
Solution. Seeking a particular solution in the form g(t) = Ae
g = 8Ae 3t and from g 00 g = e 3t we infer 8A = 1. Thus
x(t) = c1 et + c2 e
C6. Solve x00
t
+
1
e
8
3t
3t
we find
.
x = 3e2t .
Solution. Setting g(t) = Ae2t , one has g 00 g = 4Ae2t Ae2t = 3Ae2t .
Imposing that g 00 g = 3e2t one yields A = 1. Then the solution is
x(t) = c1 et + c2 e
t
+ e2t .
61
C7. Solve x00
x = te2t .
Solution. Setting g(t) = Q(t)e2t with Q = At + B we find g 00 g =
(Q + 4Q0 + 3Q)e2t . Then g 00 g = te2t yields Q00 + 4Q0 + 3Q = t, namely
4A + 3(At + B) = t. This implies 3A = 1 and 4A + 3B = 0 and hence A = 13 ,
B = 49 . Therefore g = 13 t 49 e2t and
✓
◆
1
4 2t
t
t
x = c1 e + c2 e +
t
e .
3
9
00
C8. Solve x00
3x0
x = t2 + t.
Solution. The general solution of the associated homogeneous equation
x
3x0 x = 0 is
p
p
3+ 7
3
7
2
2
x(t) = c1 e
+ c2 e
.
00
Let us try to find one solution of the non homogeneous equation in the form
g(t) = At2 + Bt + C. We need to solve g 00 3g 0 g = t2 + t. One has
2A 3(2At + B) (At2 + Bt + C) =
At2
(6A + B)t + 2A 3B
C = t2 + t.
Then A = 1, (6A + B) = 1, 2A 3B C = 0 that is A =
1 6A = 5, C = 2A 3B = 17. Thus the solution is
x(t) = c1 e
C9. Solve x00
p
3+ 7
2
+ c2 e
3
p
7
2
t2 + 5t
1, B =
17.
4x0 + 13x = 20et .
Solution. The general solution of x00 4x0 +13x = 0 is x(t) = e2t (c1 sin 3t + c2 cos 3t).
Looking for a particular solution in the form g(t) = Aet we find g 00 4g 0 +
13g = 10Aet . Then g 00 4g 0 + 13g = 20et yields A = 2 whence
x(t) = e2t (c1 sin 3t + c2 cos 3t) + 2et .
C10. Solve x00
x0
2x = 2t + et .
62
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
Solution. The general solution of x00 x0 2x = 0 is x(t) = c1 e t + c2 e2t .
Looking for a particular solution in the form g(t) = At + B + Cet we find
g 00
g0
Then g 00
A = 1,
2g = Cet
(A + Cet )
2(At + B + Cet ) =
g 0 2g = 2t + et yields 2At
2C = 1, A + 2B = 0. Thus
x(t) = c1 e
t
+ c2 e2t
2Cet
t+
1
2
2At
A
2Cet
A
2B.
2B = 2t + et whence
1 t
e.
2
C11. Solve x00 + 4x = cos t.
Solution. The general solution of x00 +4x = 0 is x(t) = c1 sin 2t+c2 cos 2t.
Looking for a particular solution in the form g(t) = A cos t we find g 00 + 4g =
A cos t + 4A cos t = 3A cos t. Therefore g 00 + 4g = cos t yields A = 13 . Then
g = 13 cos t and
1
x(t) = c1 sin 2t + c2 cos 2t + cos t.
3
C12. Solve x00 + x = sin 2t
cos 3t.
Solution. The solution has the form x(t) = c1 sin t+c2 cos t+g1 (t)+g2 (t)
where
g100 + g1 = sin 2t,
g200 + g2 = cos 3t
We look g1 , g2 in the form g1 = A sin 2t and h2 = B cos 3t we find
g1 (t) =
1
sin 2t,
3
and then
x(t) = c1 sin t + c2 cos t
g2 (t) =
1
cos 3t
8
1
1
sin 2t + cos 3t.
3
8
C13. Solve x00 + 2x0 + 2x = cos 2t.
Solution. The solution of x00 +2x0 +2x = 0 is x(t) = e t (c1 sin t + c2 cos t).
We seek a particular solution in the form g(t) = A cos 2t+B sin 2t. The equation g 00 + 2g 0 + 2g = cos 2t implies
4A cos 2t
4B sin 2t
4A sin 2t + 4B cos 2t + 2A cos 2t + 2B sin 2t =
63
= ( 2B
Then 2B
and hence
4A) sin 2t + ( 2A + 4B) cos 2t = cos 2t.
4A = 0 and
x=e
t
1
,
10
2A + 4B = 1. Solving we get A =
(c1 sin t + c2 cos t)
B=
1
5
1
1
cos 2t + sin 2t.
10
5
C14. Solve x00 + x = t sin 2t.
Solution. Set g = Q1 sin 2t + Q2 cos 2t, with Q1 = A1 t + B1 , Q2 =
A2 t + B2 . Straight calculations yield
g 00 = 4(A2
B1
A1 t) sin 2t + 4(A1
g 00 + g = (4A2
3B1
3A1 t) sin 2t + (4A1
B2
A2 t) cos 2t
whence
3B2
3A2 t) cos 2t.
If g 00 + g = t sin 2t then
4A2
Solving, A1 =
g = 13 t sin 2t
3B1
3A1 t = t,
1
, A2 = 0, B1 = 43 A2
3
4
cos 2t and hence
9
x = c1 sin 2t + c2 cos 2t
C15. Solve x00
4A1
3B2
3A2 t = 0.
= 0 and B2 =
1
t sin 2t
3
4
A
3 1
=
Then
4
cos 2t.
9
x0 = t.
Solution. The general solution is x(t) = c1 + c2 t + g(t) where g 00
We are in the resonant case. Setting g(t) = At2 + Bt we find
g 00
4
.
9
g 0 = 2A
(2At + B) = t =) A =
Thus we find
x(t) = c1 + c2 t
1 2
t
2
t.
1
,B=
2
1
g 0 = t.
64
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
C16. Solve x00
x = ekt , k 2 R.
Solution. The forcing term ekt is resonant if and only if k = ±1. If
k 6= ±1 we seek a particular solution as g = Aekt . One has g 00 g =
Ak 2 ekt Aekt = A(k 2 1)ekt . Hence g 00 g = ekt yields A(k 2 1) = 1 and
thus
1
x(t) = c1 et + c2 e t + 2
ekt , (k 6= ±1).
k
1
If k = ±1, setting g = Atekt we get g 00 g = [k 2 At+2kA]ekt Atekt = 2kAekt .
Imposing g 00 g = ekt we find 2kA = 1. Then
x(t) = c1 et + c2 e
t
x(t) = c1 et + c2 e
t
C17. Solve x00
x0
+
1 t
te ,
2
1
t e t,
2
(k = 1),
(k =
1).
2x = 3e t .
Solution. The general solution of the associated homogeneous equation
is x = c1 e t + c2 e2t . We are in the resonant case. Let us look for a particular
solution in the form z = Ate t . Then z 0 = (A At)e t , z 00 = ( 2A + At)e t
and hence
z 00
z0
2z = ( 2A + At)e
t
(A
At)e
t
2Ate
t
=
5Ae t .
It follows that z = Ate t is a solution provided 5Ae t = 3e t namely
A = 53 . In conclusion the general solution is x = c1 e t + c2 e2t 53 te t .
C18. Solve x00
3x0 + 2x = 3tet .
Solution. The general solution of the associated homogeneous equation
is x = c1 et + c2 e2t . The forcing term 3tet is resonant. Let us look for a
particular solution in the form z = (At + Bt2 )et . Since
z 0 = (A + 2Bt)et + (At + Bt2 )et = [A + (A + 2B)t + Bt2 ]et
and
z 00 = (A + 2B + 2Bt)et + [A + (A + 2B)t + Bt2 ]et
= [2A + 2B + (A + 4B)t + Bt2 ]et ,
65
then
z 00
3z 0 + 2z = [2A + 2B + (A + 4B)t + Bt2 ]et
3[A + (A + 2B)t + Bt2 ]et + 2(At + Bt2 )et
= ( A + 2B 2Bt)et .
Thus the equation z 00
3z 0 + 2z = 3tet yields,
A + 2B
whence B = 32 and A = 2B =
x = c1 et + c2 e2t
3t + 32 t2 et .
C19. Solve x00
2Bt = 3t,
3. It follows that the general solution is
4x0 + 3x = 2et
5e2t
Solution. The general solution of the associated homogeneous equation
is x = c1 et + c2 e3t . The forcing term et is resonant while e2t is not. Let us
look for a particular solution in the form z = Atet + Be2t . One finds
z 0 = (A + At)et + 2Be2t ,
z 00 = (2A + At)et + 4Be2t
and hence
z 00
4z 0 + 3z = [2A + At 4(A + At) + 3At]et + (4B
= 2Aet Be2t .
8B + 3B)e2t
Then z is a particular solution provided 2Aet Be2t = 2et 5e2t , whence
A = 1, B = 5 and z = tet + 5e2t . Hence the general solution is x =
c1 et + c2 e3t tet + 5e2t .
p
C20. Solve x00 + 2x = cos 2 t
Solution. The solution of the associated homogeneous equation is
p
p
x(t) = c1 sin 2 t + c2 cos 2 t.
p
The forcing term cos p2 t is resonant. Seeking for a particular solution in
the form g(t) = At sin 2 t we find
p
p
p
p
p
g 00 = 2 A cos 2 t + 2 A cos 2 t 2At sin 2 t
p
p
p
= 2 2 A cos 2 t 2At sin 2 t
66
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
and hence
p
p
g 00 + 2g = 2 2 A cos 2 t
2At sin
p
2 t + 2At sin
p
p
p
2 t = 2 2A cos 2 t.
p
1
From g 00 + 2g = cos 2 t it follows that A = p and thus the solution is
2 2
x(t) = c1 sin
p
2 t + c2 cos
p
p
1
2 t + p t sin 2 t.
2 2
C21. Solve x00 + 4x = sin 2t.
Solution. The solution of the associated homogeneous equation is
x(t) = c1 sin 2t + c2 cos 2t.
The forcing term sin 2t is resonant. Seeking for a particular solution in the
form g(t) = At cos 2 t we find
g(t) =
1
t cos 2t
4
and the general solution is
x(t) = c1 sin 2t + c2 cos 2t
1
t cos 2t.
4
C22. Solve x00 + x = 2 sin t + 2 cos t.
Solution. The solution is
x(t) = c1 sin t + c2 cos t + g1 (t) + g2 (t)
where g100 + g1 = 2 sin t and g200 + g2 = 2 cos t. Both are resonant. Seeking
g1 , g2 in the form g1 = at cos t and g2 = at sin t one finds
g1 (t) =
t cos t,
g2 (t) = t sin t
67
C23. Solve x00 + 9x = sin t + sin 3t.
Solution. The solution has the form
x(t) = c1 sin 3t + c2 cos 3t + g1 (t) + g2 (t),
where g1 , g2 solve, respectively
g100 + 9g1 = sin t,
g200 + 9g2 = sin 3t.
The first equation is not resonant and we find
g1 (t) =
1
sin t.
8
The second is resonant, and we find
1
t cos 3t.
6
g2 (t) =
C24. Solve the boundary value problem x00
x = t, x(0) = x(1) = 0.
Solution. The general solution is x(t) = c1 et + c2 e t t. The boundary
condition x(0) = 0 yields c1 + c2 = 0 and hence x(t) = c1 et c1 e t t. For
t = 1 one finds x(1) = c1 e c1 e 1 1 = 0 which gives
c1 =
1
e
1
e
=
e
e2
1
Thus the unique solution of the boundary value problem is
e
x(t) = 2
et e t
t.
e
1
C25. Find k such that the solution of x00 + 4x0 + x = k, x(0) = 0, x0 (0) = 0
tends to 1 as t ! +1.
Solution. The general solution is
x(t) = c1 e↵t + c2 e t + k,
where ↵ =
2+
Imposing x(0) = x0 (0) = 0 we deduce the system
⇢
c1 + c2 + k = 0
↵c1 + c2 = 0
p
3,
=
2
p
3.
68
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
which has the unique solution c1 =
x(t) =
k
k
↵
< 0, then x(t) !
Since ↵ > 0 and
↵k
e↵t +
↵
↵k
, c2 =
↵
↵
. Thus we find
e t + k.
1 as t ! +1 if and only if k > 0.
C26. Show that if 6= 0 and h(t) > 0 then any (possible) nontrivial solution
2
of the boundary value problem x00
x = h(t), x(a) = x(b) = 0 has to
change sign in (a, b).
Solution. Multiplying the equation by the solution x(t) and integrating
in [a, b] we find
Z
b
00
2
x (t)x(t)dt
a
Z
b
2
x (t)dt =
a
Z
b
h(t)x(t)dt.
a
Integrating by parts and taking into account that x(a) = x(b) = 0, one infers
Z
Then
Z
b
b
Z
00
x (t)x(t)dt =
a
02
x (t)dt
a
2
Z
b
x02 (t)dt.
a
b
2
x (t)dt =
a
Z
b
h(t)x(t)dt < 0.
a
Since h(t) > 0 in (a, b), it follows that x(t) has to change sign.
C27. Show that for all a 6= 0 the boundary value problem x00
x(0) = x(a) = 0, has a unique solution.
Solution. The general solution is
x(t) = c1 e
p
2t
+ c2 e
p
2t
2et .
The conditions x(0) = 0, x(a) = 0 yield
⇢
c1 +
c
p 2
2a
c1 e
+ c2 e
p
2a
=2
= 2ea
2x = 2et ,
69
For a 6= 0, the determinant
1
e
p
2a
e
1p
2a
=e
p
2a
e
p
2a
6= 0
Hence the system has a unique solution c1 , c2 .
D1. Show that
t5 + 1
x + 4
x=0
t +5
00
is an oscillatory equation.
Solution. There exists T > 0 such that
t5 + 1
> 1,
t4 + 5
for all t
T.
Then by the Sturm comparison theorem the given equation is oscillatory because so is x00 + x = 0.
D2. Which one of the following two equations has solutions that oscillate
more rapidly?
p
x00 + t6 + 3t5 + 1 x = 0,
x00 + 2t3 x = 0.
Solution. First we note that comparing these two equations to x00 + x = 0
implies that both are oscillatory. Since
2t3
=2
t!1
t6 + 3t5 + 1
p
there exists a number T such that 2t3 > t6 + 3t5 + 1 for t > T .
Let x1 and x2 be nontrivial solutions of the first and second equations
respectively such that x1 (a) = x2 (a) = 0, where a > T is some number. We
need to show that x2 oscillates faster, that is if x1 (a) = x1 (b) = 0, b > a,
then there exists a number c between a and b such that x2 (c) = 0. But this
follows from the Sturm Comparison Theorem.
lim p
D3. Explain why no nontrivial solution of x00 (t) + p(t)x(t) = 0 can vanish at
each of the numbers 0, 1, 1/2, 1/3, ...1/n....
70
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
Solution. Otherwise, as in the proof of Theorem 5.7.12, one shows that
x(0) = x0 (0) = 0. Thus, by uniqueness, one has that x(t) ⌘ 0, a contradiction.
D4. Consider the boundary value problem
x00
p(t)x = q(t), x(a) = x(b) = 0.
Show that if p(t) and q(t) are continuous, with p(t) > 0, on the interval [a, b],
then there is a unique solution of this boundary value problem.
Solution. Let x1 , x2 be two the linearly independent solutions of x00 p(t)x =
0 satisfying the initial conditions x1 (a) = 1, x01 (a) = 0 and x2 (a) = 0, x02 (a) =
1. Let y = c1 x1 + c2 x2 + xp be the general solution of x00 px = q, where xp
is some particular solution of this equation. Now we show that c1 , c2 can be
chosen so that y(a) = y(b) = 0. But there exists a unique set of constants
c1 , c2 satisfying the algebraic system
⇢
c1 x1 (a) + c2 x2 (a) + xp (a) = 0
c1 x1 (b) + c2 x2 (b) + xp (b) = 0
since the coefficient determintant is equal to x1 (a)x2 (b) x1 (b)x2 (a) = x2 (b) 6=
0 (by Proposition 5.7.9, x2 (t) cannot have more than one zero).
D5. Show that, assuming that p0 (t) and q(t) are continuous, the substitution
x(t) = y(t)e
1
2
R
p(t)dt
transforms the equation
x00 + p(t)x0 + q(t)x = 0
into the form x00 (t) + P (t)x(t) = 0.
1
Solution. Since y = xe 2
1
R
R
p(t)dt
it follows
1
R
1
R
y 0 = x0 e 2 p(t)dt + 12 xpe 2 p(t)dt
= x0 + 12 xp e 2 p(t)dtR,
R
1
1
y 00 = x00 + 12 x0 p + 12 xp0 e 2 p(t)dt + 12 p x0 R+ 12 xp e 2 p(t)dt
1
= x00 + 12 x0 p + 12 xp0 + 12 x0 p + 14R xp2 e 2 p(t)dt
1
= x00 + x0 p + 12 xp0 + 14 xp2 e 2 p(t)dt .
71
If x00 + px0 =
qx we infer
1
1
Since y = xe 2
R
p(t)dt
R
qx + 12 xp0 + 14 xp2 Re 2 p(t)dt
1
q + 12 p0 + 14 p2 xe 2 p(t)dt .
y 00 =
=
it follows
00
y =
✓
◆
1 0 1 2
q+ p + p y
2
4
which is an equation of the form x00 + P (t)x = 0.
D6. Determine the oscillation of
x00 + x0 + x = 0
in two ways.
a) by using the Sturm Comparison theorem,
b) by solving the equation explicitlly.
Solution.
a) We use the previous exercise with p = q = 1. Then y =
R
1
1
dt
xe 2
= xe 2 t satisfies y 00 + 34 y = 0, which is an oscillatory equation. Of
1
course, if y is oscillatory, so is x = ye 2 t .
b) The
solutions of the characteristic equation m2p+ m + 1 =
0 are
p
p
1
1±i 3
3
3
m=
and hence the general solution is x = (c1 sin 2 t+c2 cos 2 t)e 2 t
2
which is oscillatory.
D7. Determine the oscillation of
x00
1 0
tx + x = 0.
4
Solution. Using exercise D5, with p = 14 t and q = 1, we find that
1 2
1 2
1 2
y = xe 16 t satisfies y 00 =
1 18 + 64
t y namely y 00 + 98 64
t y = 0.
This equation is nonoscillatory by Proposition 5.7.9.
D8. Let u00 + p1 (t)u = 0 and v 00 + p2 (t)v = 0, with v(t) 6= 0 in [a, b].
(a) Prove the Picone Identity
⇣u
⌘0
⇣
u ⌘2
(u0 v uv 0 ) = (p2 p1 )u2 + u0 v 0
v
v
72
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
(b) Use this to prove the Sturm comparison theorem.
Solution. (a) One has
⇣u
v
(u0 v
uv 0 )
⌘0
⇣ u ⌘0
u
uv 0 ) + (u0 v uv 0 )0
v
v
u0 v uv 0 0
u
=
(u v uv 0 ) + (u00 v + u0 v 0
2
v
v
0
0 2
00
(u v uv ) + uv(u v uv 00 )
=
v2
=
(u0 v
Using the equations, i.e. u00 =
⇣u
v
(u0 v
uv 0 )
⌘0
(u0 v
p1 u and v 00 =
u0 v 0
p2 v, we infer
uv 0 )2
uv( p1 uv + p2 uv)
v2
(u0 v uv 0 )2 + (p2 p1 )u2 v 2
=
v2
⇣
(u0 v uv 0 )2
2
= (p2 p1 )u2 +
=
(p
p
)u
+
u0
2
1
v2
=
uv 00 )
v 00
u ⌘2
v
(b) Let a < b be two consecutive zeros of u. Suppose that u > 0 in (a, b)
(the case that u < 0 is similar). If p2 > p1 the right hand side of the Picone
identity is positive and thus
✓
u(t) 0
(u (t)v(t)
v(t)
0
u(t)v (t)
◆0
> 0,
8 t 2 [a, b].
Then the function on the left hand side is increasing and thus
u(b) 0
(u (b)v(b)
v(b)
u(b)v 0 (b)) >
u(a) 0
(u (a)v(a)
v(a)
u(a)v 0 (a)).
But u(a) = u(b) = 0 and hence both the right and the left hand sides of the
preceding equality are zero and we find a contradiction.
D9. Let u00 + p1 (t)u = 0 and v 00 + p2 (t)v = 0 and that p2 (t) > p1 (t) in (a, b).
Suppose that u(a) = v(a) = 0, u0 (a) = v 0 (a) = ↵ > 0. Show that there exists
✏ > 0 such that v(t) > u(t) in (a, a + ✏)
73
Solution. First of all, from the initial conditions it follows that there exists
✏ > 0 such that both u(t) > 0 and v(t) > 0 in (a, a+✏). Consider the function
(t) =
v(t)
,
u(t)
t 2 (a, a + ✏).
Notice that, by the Hopital rule
lim
t!a+
v 0 (t)
v 0 (a)
=
= 1,
t!a+ u0 (t)
u0 (a)
(t) = lim
and thus
is continuous at t = a with (a) = 1. Multiplying the first
equation by v and the second by u and subtracting, we get
u00 v
Then
0
=
uv 00 = ( p1 + p2 )uv > 0,
v0u
t 2 (a, a + ✏).
u0 v
> 0 in (a, a + ✏) and hence us strictly increasing
u2
in (a, a + ✏). Therefore (t) > (a) = 1 in (a, a + ✏) which implies that
v(t) > u(t) in (a, a + ✏).
x0
, x(1) = 0, x0 (1) = 1.
t
x0
Solution. A particular solution of x00 =
is x0 = 0, namely x(t) constant,
t
which does not satisfy the initial conditions. Setting z = x0 we find the first
order separable equation z 0 = z/t. Either z = 0 or z does not change sign.
Since z(0) = x0 (0) = 1 it follows that z(t) 6= 0 for all t and hence, dividing
by z and integrating,
D10. Solve the initial value problem x00 =
dz
dt
=
z
t
that is
ln |z| = ln |t| + c
Since z(1) = x0 (1) = 1 we find c = 0. Then x0 (t) = t and hence, integrating
and taking into account that x(1) = 0, we have
Z
t
0
x (t)dt =
1
Z
t
1
tdt that is x(t) = 12 t2
1
2
.
74
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
D11. Solve x00 = 2x0 (x
1), x(0) = 0, x0 (0) = 1.
Solution. A particular solution is x = constant which does not satisfy the
dz
initial condition x0 (0) = 1. Setting z = x0 one has x00 = z
and hence
dx
z
dz
= 2z(x
dx
1),
z(0) =
1.
dz
= 2(x 1) whereby z = (x 1)2 + c1 . From
dx
x = 0, z(0) = 1 it follows c1 = 0. Then x0 = (x 1)2 , with x(0) = 0 and,
integrating,
Z
dx
= t + c2
(x 1)2
Dividing by z 6= 0, we find
whereby
1
= t + c2 .
x 1
From x(0) = 0 we infer that c2 = 1. Therefore
x=1
1
t
=
.
t+1
t+1
D12. Solve x00 = 2x03 x.
Solution. Setting z = x0 we get
z
dz
= 2z 3 x.
dx
One solution is z = 0 which yields x(t) = const. Otherwise, we find
which is separable. Integrating we obtain
Z
Z
dz
= 2xdx
z2
whereby
1
= x2 + c1
z
Then
z = x0 =
x2
1
+ c1
namely
(x2 + c1 )x0 =
1.
dz
dx
= 2z 2 x
75
Integrating we find 13 x3 + c1 x =
form.
D13. Solve xx00
2x02
t + c2 , which gives the solutions in implicit
x2 = 0.
Solution. The equation is homogeneous of degree 2 with respect to x, x0 , x00 .
Setting x0 = xz one has x00 = xz 2 +xz 0 and hence x(xz 2 +xz 0 ) 2x2 z 2 x2 = 0,
that is x2 z 2 + x2 z 0 2x2 z 2 x2 = 0 or else x2 (z 0 z 2 1) = 0. Let us solve
z 0 z 2 1 = 0, which is a separable equation. Integrating, one finds
Z
Z
dz
= dt,
z2 + 1
which implies arctan z = t + c1 that is
z = tan(t + c1 ),
⇡
⇡
< t + c1 < .
2
2
From x0 = xz we infer x0 = x tan(t + c1 ). By uniqueness, either x(t) ⌘ 0 or
x(t) never vanishes. Integrating
Z
Z
dx
= tan(t + c1 )dt
x
which yields ln |x(t)| =
find
|x(t)| =
ln | cos(t + c1 )| + c2 . Solving with respect to x we
e c2
⇡
⇡
,
< t + c1 < .
| cos(t + c1 )|
2
2
To this family we have to add the trivial solution x(t) ⌘ 0.
D14. Solve (a) xx00
x02 + et x2 = 0, x(0) = 1, x0 (0) =
00
02
t 2
xx
x + e x = 0, x(0) = 1, x0 (0) = 1
1, and (b)
Solution. (a) Setting x0 = xz one has x00 = xz 2 + xz 0 which implies x2 z 0 +
et x2 = 0 that is x2 (z 0 + et ) = 0. Solving z 0 + et = 0 we find by integration
z(t) = et + c1 . The conditions x(0) = 1, x0 (0) = 1, yield z(0) = 1 and
hence c1 = 0. Thus z(t) = et , that is x0 = xz = xet . Since x(t) cannot
change sign and x(0) = 1, then x(t) > 0 and we have ln x(t) = et + c2 .
From x(0) = 1 we infer c2 = 1 and thus the problem has a unique solution
given implicitly by ln x(t) = 1 et namely by
x = e1
et
.
76
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
Another solution of x2 (z 0 + et ) = 0 is given by x(t) ⌘ 0 which, however, does
not satisfy the initial conditions.
(b) As before, z(t) = et +c1 . Now the conditions x(0) = 1, x0 (0) = 1,
yield z(0) = 1 and hence c1 = 2. Thus z(t) = 2 et , that is x0 = xz = x(2 et )
and ln |x(t)| = 2t et + c2 . For t = 0 one has ln |x(0)| = 1 + c2 and since
x(0) = 1, c2 = 1. Therefore ln |x(t)| = 1 + 2t et . Also here x(t) ⌘ 0 solves
the equation xx00 x02 + et x2 = 0 but not the initial conditions.
D15. Solve the Euler equation t2 x00
2x = 0, t > 0.
Solution. We set s = ln t, t > 0, finding
d2 x
ds2
dx
ds
2x = 0.
The characteristic equation of this linear di↵erential equation is m2 m 2 =
0 whose roots are m1 = 1, m2 = 2 and thus x = c1 e s + c2 s2s . Since es = t
we find
1
x(t) = c1 + c2 t2 .
t
D16. Solve t2 x00 + atx0 + x = 0, t > 0.
Solution. Setting s = ln t, t > 0, we find
d2 x
+ (a
ds2
1)
dx
+ x = 0.
ds
The characteristic equation is m2 + (a 1)m + b = 0 whose roots are
p
p
(a 1) ± (a 1)2 4(a 1)
1 a ± a2 6a + 5
m1,2 =
=
2
2
There are 3 cases.
(i) If a < 1 or a > 5 one has a2
distinct. In this case
x(s) = c1 em1 s + c2 em2 s
6a + 5 > 0 and the roots are real and
whereby x(t) = c1 tm1 + c2 tm2 .
77
(ii) If a = 1, resp. a = 5, then there is a double root m1 = 0, resp.
m2 = 2. Then if m1 = 0 we find
x(s) = c1 + c2 s whereby x(t) = c1 + c2 ln t,
while, if m2 =
2,
x(s) = c1 e
2s
+ c2 se
2s
whereby x(t) = c1 t
2
(iii) If 1 < a < 5 the roots are complex conjugate
p
↵ = 12
a2 + 6a 5. Then
x(s) = e
1 a
s
2
2
+ c2 t
1
a
2
ln t.
± i↵, where
(c1 sin ↵s + c2 cos ↵s)
which yields
x(t) = t
D17. Solve t2 x00
tx0
1 a
2
(c1 sin(↵ ln t) + c2 cos(↵ ln t)).
3x = 0, x(1) = 0, x0 (1) = 1, t > 0.
Solution. If t = es , we find
d2 x
ds2
2
dx
ds
3x = 0
1
which yields x = c1 e s + c2 s3s and hence x(t) = c1 + c2 t3 . Since x0 =
t
1
2
c1 2 + 3c2 t , the initial conditions imply 0 = c1 + c2 , 1 = c1 + 3c2
t
1
1
namely c1 =
, c2 = . In conclusion
4
4
x(t) =
1
t3
t4 1
+ =
.
4t
4
4t
D18. Solve the nonhomogeneous Euler equation t2 x00 + tx0 + x = t, t > 0.
78
CHAPTER 5. SOLUTIONS TO EXERCISES OF CHAPTER 5
Solution. Setting es = t that is s = ln t, t > 0, we find
d2 x
+ x = es .
ds2
The general integral of ẍ + x = 0 is x = c1 sin s + c2 cos s. A particular
solution can be found in the form h(s) = aes . From ḧ + h = es we find a = 12
Thus x = c1 sin s + c2 cos s + 12 es Substituting s = ln t we get
1
x(t) = c1 sin(ln t) + c2 cos(ln t) + t,
2
D19. Solve t2 x00 + 3tx0
3x = t2 ,
t>0
t > 0.
Solution. If t = es , the corresponding equation for x(s) is
d2 x
dx
+
2
ds2
ds
3x = e2s
The characteristic equation of the associated homogeneous equation is m2 +
2m 3 = 0 whose solutions are m1 = 1, m2 = 3. A particular solution of
the non homogeneous equation is 15 e2s and thus the general solution is
x(s) = c1 es + c2 e
Since t = es we find
x(t) = c1 t +
D20. Show that a solution of x00
3.
3s
+
1 2s
e .
5
c2 1 2
+ t .
t3 5
tx0 + 3x = 0 is a polynomial P of degree
Solution. Set P (t) = ao + a1 t + a2 t2 + a3 t3 and seek ai in such a way that
P 00 = tP 0 3P . One finds
2a2 + 6a3 t = t(a1 + 2a2 t + 3a3 t2 )
3(a0 + a1 t + a2 t2 + a3 t3 )
namely
2a2 + 6a3 t =
3a0
2a1 t
a2 t2 .
From the identity principle of polynomials we find 2a2 = 3a0 , 6a3 =
0. Thus a0 = a2 = 0, a1 = 3a3 and hence P (t) = 3a3 t + a3 t3 .
2a1 , a2 =
Chapter 6
Solutions to exercises of
Chapter 6
1. Find the general solution if 2x000 = 0.
Solution. The characteristic equation is 2m3 = 0 whose root is m = 0
with multiplicity 3. Thus x = c1 + c2 t + c3 t2 .
2. Find the general solution of x000
x0 = 0.
Solution. The characteristic equation is m3
m = 0, 1, 1. Thus x = c1 + c2 e t + c3 et .
3. Find the general solution of x000 + 5x0
m = 0 whose roots are
6x = 0.
Solution. One root of the characteristic equation m3 + 5m 6 = 0 is
m = 1. Then m3p+ 5m 6 = (m 1)(m2 + m + 6) . Thus the other
1 ± i 23
roots are
and the general solution is
2
"
!
!#
p
p
1
23
23
x = c1 et + e 2 t c1 sin
t + c3 cos
t
.
2
2
4. Find the general solution of x000
4x00 + x0
4x = 0.
Solution. A root of m3 4m2 + m 4 = 0 is m = 4. Dividing by
m 4 one finds m3 4m2 + m 4 = (m 4)(m2 + 1). Then the roots
are m = 4, ±i and the general solution is x = c1 e4t + c2 sin t + c3 cos t.
79
80
CHAPTER 6. SOLUTIONS TO EXERCISES OF CHAPTER 6
5. Find the general solution of x000
3x00 + 4x = 0.
Solution. One has m3 3m2 +4 = (m+1)(m2 4m+4) and hence the
roots are m = 1 and m = 2, double. Thus x = c1 e t + c2 e2t + c3 te2t .
6. Find the solution to the initial value problem x000 + 4x0 = 0, x(0) =
1, x0 (0) = 1, x00 (0) = 2.
Solution. The general solution is
x = c1 + c2 sin 2t + c3 cos 2t.
Substituting the initial values and solving for the constants, we obtain
c1 + c3 = 1, 2c2 = 1 and 4c3 = 2. Then c1 = 32 , c2 = c3 = 12 .
Therefore the desired solution is x(t) = 32 12 sin 2t 12 cos 2t.
7. Find the solution of x000 x0 = 0 satisfying x(0) = 1 and limt!+1 x(t) =
0.
Solution. The roots of m3 m = 0 are m = 0, ±1 and hence the
general solution is x(t) = c1 +c2 et +c3 e t . The condition limt!+1 x(t) =
0 implies that c1 = c2 = 0. Moreover, x(0) = 1 yields c3 = 1. Thus the
solution we are looking for is x(t) = e t .
8. Find the solution of x000 x0 = 0, x(0) = 0, x0 (0) = 0, x00 (0) = 1.
Solution. We know that the general solution is x = c1 +c2 et +c3 e t .The
initial conditions give rise to the system
8
< c1 + c2 + c3 = 0
c2 c3
=0
:
c2 + c3
=1
whose solutions are c1 =
1
(e t + et ) = 1 + cosh t.
2
1 and c2 = c3 =
1
.
2
Thus x =
1+
9. Solve the initial value problem
x000 + x00
2x = 0, x(0) = 0, x0 (0) = 0, x00 (0) = 1.
Solution. Solving the characteristic equation m3 + m2 2 = (m
1)(m2 + 2m + 2) = 0, we obtain m = 1, 1 ± i. Therefore, the general
solution is x(t) = c1 et + e t (c2 sin t + c3 cos t). Setting x(0) = x0 (0) =
81
0, x00 (0) = 1, we obtain the algebraic equations c1 +c3 = 0, c1 +c2 c3 =
0, c1 2c2 = 1. Solving these equations, we obtain c1 = 51 , c2 =
2
, c3 = 15 . Therefore the desired solution is
5
1
x(t) = et
5
e
t
✓
◆
2
1
sin t + cos t .
5
5
10. Show that there exists a solution of x000 + ax00 + bx0 + cx = 0 such that
limt!+1 x(t) = 0, provided c > 0.
Solution. Instead of solving the characteristic equation C(m) = m3 +
am2 + bm + c = 0, let us notice that the function C(m) satisfies C(0) =
c > 0 and limm! 1 C(m) = 1. Thus C has at least one negative
zero m = a < 0 which gives rise to the solution x = e at that satisfies
the required limit condition.
11. Show that for 0 < k < 2 the equation x000 3x0 + kx = 0 has a unique
solution such that x(0) = 1 and limt!+1 x = 0.
Solution. Consider the characteristic equation C(m) = m3 3m+k =
0. Since C 0 (m) = 3(m2 1) = 0 yields m = ±1, then C has a maximum
at m = 1 and a minimum at m = 1. Since C(1) = k 2 < 0 and
C(m) ! 1 as m ! 1, there is a root to the right of m = 1. Also, since
C(0) = k > 0, there is a root between m = 0 and m = 1. Since C(0) >
0 and C(m) ! 1 as m ! 1, there is a root to the left of m = 0.
Therefore there is exactly one negative root m1 and two positive roots
m2 , m3 . The general solution is x = c1 em1 t + c2 em2 t + c3 em3 t . Hence the
condition x(0) = 1 and limt!+1 x(t) = 0 implies c1 = 1, c2 = c3 = 0.
Thus x = em1 t .
12. Find the general solution of x0000
6x00 + 5x = 0.
Solution. The characteristic equation is m4 6m2 +5 =p0. Solving this
bi-quadratic
p
p equation we find 4 real solutions m = ± 3 p± 2, namely
m = ±1, ± 5. The general solution is x = c1 et +c2 e t +c3 e 5t +c4 e 5t .
13. Find the solution of x0000
x = 0, x(0) = 1, x0 (0) = x00 (0) = x000 (0) = 0.
Solution. The characteristic equation is m4 1 = 0 whose roots are
m = ±1 and m = ±i. The general solution is x = c1 et + c2 e t +
82
CHAPTER 6. SOLUTIONS TO EXERCISES OF CHAPTER 6
c3 sin t + c4 cos t. The initial conditions
8
c1 + c2 + c4
>
>
<
c1 c2 + c3
c1 + c2 c4
>
>
:
c1 c2 c3
give rise to the system
=1
=0
=0
= 0.
Solving the above algebraic system, we find c1 = c2 = 14 , c3 = 0, c4 = 12 .
Hence the solution is x(t) = 14 et + 14 e t + 12 cos t.
14. Find the solutions of x0000
x00 = 0, x(0) = 1, x00 (0) = 0.
Solution. The roots of the characteristic equation m4 m2 = 0 are
m = 0 (double) and m = ±1. The general solution is x = c1 + c2 t +
c3 et + c4 e t . The conditions x(0) = 1, x00 (0) = 0 yield c1 + c3 + c4 =
1, c3 + c4 = 0, that is c1 = 1 and c3 = c4 . Thus the solutions are
x(t) = 1 + c2 t + c3 et c3 e t , depending on the two real parameters
c 2 , c3 .
15. Find the solution x(t) of x0000
and x(0) = 0, x0 (0) = 1.
4x00 + x = 0 such that limt!+1 x(t) = 0
Solution. The bi-quadratic equation m4 4m2 +1 = 0 has 4 real roots
±a, ±b where
q
q
p
p
a = 2 + 3,
b= 2
3.
The general solution is x = c1 e at +c2 e bt +c3 eat +c4 ebt . If limt!+1 x(t) =
0 then c3 = c4 = 0. We now impose the initial conditions x(0) =
0, x0 (0) = 1, finding the system
⇢
1
1
c1 + c2
=0
yielding c1 =
, c2 =
ac1 bc2 = 1
a b
a b
Thus x(t) =
e
a
bt
b
e
a
at
b
=
e
bt
a
e
b
at
.
16. Show that the only solution of x0000 8x000 + 23x00 28x0 + 12x = 0, such
that limt!+1 x(t) = 0, is the trivial solution x(t) ⌘ 0.
Solution. The characteristic equation is m4 8m3 +23m2 28m+12 =
0. Looking for roots that are divisors of 12, we find m = 1, m = 2
(double) and m = 3. Thus the general solution is x = c1 et + c2 e2t +
c3 te2t + c4 e3t . Clearly, limt!+1 x(t) = 0 implies c1 = c2 = c3 = c4 = 0.
83
17. Show that x0000 + 2x00
x(0) = 1, x0 (0) = 1.
4x = 0 has one periodic solution such that
Solution. Solving the characteristic equation m4 +3m2 4 = 0 we find
3±5
m2 =
whereby m2 = 1 and m2 = 4 and finally m = ±1 and
2
m = ±2i. The periodic solutions are x(t) = c1 sin 2t + c2 cos 2t. The
condition x(0) = 1 yields c2 = 1. The condition x0 (0) = 1 gives 2c1 = 1.
Thus the periodic solution we are looking for is x(t) = 12 sin 2t + cos 2t.
18. Find the general solution of x(5)
x0 = 0.
Solution. Factoring m5 m = 0 we find m(m4 1) = m(m2 1)(m2 +
1) = 0 and hence m = 0, m = ±1 and m = ±i. Hence the general
solution is x = c1 + c2 et + c3 e t + c4 sin t + c5 cos t.
19. Find the general solution of x(5) + x0000
5
4
x0
x = 0.
4
Solution. Since m +m m 1 = m (m+1) (m+1) = (m+1)(m4
1) = (m + 1)(m2 1)(m2 + 1) = (m + 1)2 (m 1)(m2 + 1), it follows that
the roots of the characteristic equation are m = 1 (double), m = 1,
m = ±i. The general solution is x = c1 e t + c2 te t + c3 et + c4 sin t +
c5 cos t.
20. Show that x(5) +x = 0 has at least one solution such that limt!+1 x(t) =
0.
Solution. The characteristic equation m5 + 1 = 0 has the negative
root m = 1. The function x(t) = e t satisfies the limit requirement.
21. Find the general solution of x(6)
x00 = 0.
Solution. Since m6 m2 = m2 (m4 1), the roots of the characteristic
equation are m = 0 (double), m = ±1, m = ±i. The general solution
is x = c1 + c2 t + c3 et + c4 e t + c5 sin t + c6 cos t.
22. Find the general solution of x(6)
64x = 0.
Solution. In order to solve the characteristic equation m6 64 = 0 ,
we note that m6 64 = (m3 8)(m3 + 8) = (m p 2)(m2 +
p2m + 4)(m +
2)(m2 2m + 4) = 0, yielding m = 2, 2, 1 ± i 3, 1 ± i 3. Therefore,
the general solution is
p
p
p
p
x(t) = c1 e2t +c2 e 2t +e t (c3 sin 3 t+c4 cos 3 t)+et (c5 sin 3 t+c6 cos 3 t).
84
CHAPTER 6. SOLUTIONS TO EXERCISES OF CHAPTER 6
23. Find the general solution of x0000 + 3x000 + 2x00 = et
a) by the method of Variation of parameters.
b) by the method of Undetermined Coefficients.
Solution. First, in order to find the general solution of the homogeneous equation x0000 + 3x000 + 2x00 = 0, we find the roots of the characteristic equation m4 + 3m3 + 2m2 = 0 which has roots m = 0 (double
root), 1, 2. Thus the general solution of the homogeneous equation
is
xh (t) = c1 + c2 t + c3 e t + c4 e 2t .
a) Variation of Parameters. In order to find a particular solution of
the nonhomogeneous equation, we let xp = v1 + v2 t + v3 e t + v4 e 2t ,
and we determine v1 , v2 , v3 , v4 from the following equations:
v10 + v20 t + v30 e t + v40 e 2t = 0,
v10 · 0 + v20 · 0 + v30 · e t + 4v40 · e
v10 · 0 + v20 · 0 v30 · e t 8v40 · e
2t
2t
= 0,
= et .
We notice that the last two equations have two unknowns. So, we
solve them for v30 and v40 , obtaining v30 = e2t , v40 = 41 e3t from which we
obtain v3 = 12 e2t , v4 = 121 e3t . Now we substitute these values of v30 and
v40 into the second equation and solve for v20 obtaining v20 = 12 et from
which we find v2 = 12 et . Finally, we substitute the values of v20 , v30 , v40
into the first equation and solve for v10 , obtaining v10 = 12 tet 34 et ,
which yields v1 = 12 tet 14 et . Therefore,
xp (t) =
1 t
te
2
1 t 1 t 1 t
1 t 1 t
e + te + e +
e = e
4
2
2
12
6
and the general solution of the given nonhomogeneous equation is
x(t) = c1 + c2 t + c3 e
t
+ c4 e
2t
1
+ et .
6
b) Undetermined Coefficients. It seems reasonable that xp may be of
the form xp = Aet . Upon substitution we find Aet + 3Aet + 2Aet = et ,
which immediately yields A = 16 and hence xp = 16 et and the general
solution to the given problem is the same as in part a).
24. Find the general solution of x000 + 4x0 = sec 2t.
85
Solution. In this case, it is not clear that the method of Undetermined
Coefficients would be easy or convenient. So, we opt for the sure way
to solve it, which is the method of Variation of Parameters.
The characteristic equation m3 +4m = 0 has roots m = 0, ±2i. So, the
general solution of x000 + 4x0 = 0 is x(t) = c1 + c2 sin 2t + c3 cos 2t. Now
we find v1 , v2 , v3 such that xp = v1 + v2 sin 2t + v3 cos 2t is a solution
of x00 + 4x0 = sec 2t. In order to do so, we solve the system
8
< v10 + v20 sin 2t + v30 cos 2t = 0,
v10 · 0 + 2v20 cos 2t 2v30 sin 2t = 0,
:
v10 · 0 4v20 sin 2t 4v30 cos 2t = sec 2t.
Using Cramer’s Rule, we find that v10 = 14 sec 2t, v20 = 14 sin 2t sec 2t,
v30 = 14 . Now we integrate and obtain v1 = 18 ln | sec 2t + tan 2t|,
v2 = 18 ln | cos 2t|, v3 = 14 t. Therefore, the general solution of the
given nonhomogeneous equation is
x(t) = c1 + c2 sin 2t + c3 cos 2t
1
1
+ ln | sec 2t + tan 2t| + ln | cos 2t| sin 2t
8
8
25. Solve x000
1
t cos 2t.
4
x00 = 1.
Solution. The roots of the characteristic equation m3 m2 = 0 are
m = 0 (double) and m = 1 and hence general solution is x(t) =
c1 + c2 t + c3 et + xp , where xp solves x000
x00p = 1. Taking xp = ↵t2 we
p
1 2
find 2↵ = 1 namely xp = 2 t , and hence x(t) = c1 + c2 t + c3 et 12 t2 .
26. Solve x000
x0 = t, x(0) = x0 (0) = x00 (0) = 0.
Solution. The roots of the characteristic equation m3 m = 0 are m =
0 and m = ±1 and hence general solution is x(t) = c1 +c2 et +c3 e t +xp ,
where xp solves x000
x0p = t. Taking xp = ↵t2 we find 2↵t = t namely
p
↵ = 12 and xp = 12 t2 . Hence x(t) = c1 + c2 et + c3 e t 12 t2 . The
initial conditions yield c1 + c2 + c3 = 0, c2 c3 = 0, c2 + c3 1 = 0,
whereby c1 = 1, c2 = c3 = 12 . Then x(t) = 1 + 12 et + 12 e t 12 t2 .
27. Solve x0000 + x000 = t, x(0) = x0 (0) = x00 (0) = 0.
Solution. The roots of the characteristic equation m4 + m3 = 0 are
m = 0 (triple) and m = 1 and hence general solution is x(t) =
000
c1 + c2 t + c3 t2 + c4 e t + xp , where xp solves x0000
p + xp = t.
86
CHAPTER 6. SOLUTIONS TO EXERCISES OF CHAPTER 6
Taking xp = at4 +bt3 we find 24a+24at+6b = t, whereby a = 1/24, b =
1 4
1 3
1/6. Thus x(t) = c1 + c2 t + c3 t2 + c4 e t + 24
t
t.
6
The initial conditions yield c1 + c4 = 0, c2 c4 = 0 and 2c3 + c4 = 0.
Then c1 = c4 , c2 = c4 , c3 = 12 c4 and the solutions, depending on a
1 4 1 3
real parameter c4 , are given by x(t) = c4 +c4 t 12 c4 t2 +c4 e t + 24
t 6t .
28. Explain whether the functions
5, t, t2 , t3 , sin t, 3
t2 , cos t, et , e
t
are linearly dependent or independent.
Solution. A direct approach dealing with all of these functions can
be quite cumbersome. We recall that if a subset of these functions
consists of functions that are linearly dependent, then all of them are
linearly dependent. Looking for such a subset, the logical choice is
5, t2 , 3 t2 . It is easy to determine nonzero constants c1 , c2 , c3 such
that 5c1 + c2 t2 + c3 (3 t2 ) = 0. We notice that 5c1 + c2 t2 + c3 (3 t2 ) =
5c1 + 3c3 + (c2 c3 )t2 = 0. Let c1 = 1, c3 = 35 = c2 .
29. Explain why x(t) = sin t5 cannot be a solution of a fourth order linear
homogeneous equation.
Solution. Clearly, x(0) = x0 (0) = x00 (0) = x000 (0) = 0. Therefore, by
uniqueness, we must have x(t) ⌘ 0, which is impossible.
30. Evaluate W (t, t2 , t3 , sin t, cos t, t4 , et , e t , t4
t2 ).
Solution. Similar to Exercise 28, c1 (t4 t2 ) + c2 t2 + c3 t4 ⌘ 0 if we let
c1 = c2 = 1, c3 = 1. Therefore these three functions, and hence all the
functions whose Wronskian we wish to evaluate, are linearly dependent,
which means that W (t, t2 , t3 , sin t, cos t, t4 , et , e t , t4 t2 ) ⌘ 0.
31. Consider x0000 3x000 + 2x0 5x = 0. If x1 , x2 , x3 , x4 are solutions and
W (x1 , x2 , x3 , x4 )(0) = 5, find W (x1 , x2 , x3 , x4 )(6).
Solution. By Abel’s Theorem, W (t) = ce3t . Hence we have W (0) = c.
Since we are given W (0) = 5, then c = 5. Therefore, W (6) = 5e18 .
32. Explain why et , sin t, t cannot be solutions of a third order homogeneous equation with continuous coefficients.
87
Solution. Calculating the Wronskian, we find that W (t) = et ( t sin t
t cos t + 2 sin t). We note that W (0) = 0 while W (⇡) = ⇡e⇡ . This contradicts Abel’s Theorem which implies that the Wronskian of solutions
is either always zero or never zero.
33. Solve t3 x000 + 4t2 x00 + 3tx0 + x = 0, t > 0.
Solution. This is an Euler equation and can be changed to one with
constant coefficient by substituting t = es or equivalently, s = ln t.
Then by using the Chain Rule we calculate the derivatives with respect
to the variable s. Let ẋ = dx
. Then we have
ds
✓ ...
◆
✓
◆
✓ ◆
x 3ẍ + 2ẋ)
ẍ ẋ
ẋ
3
2
t
+ 4t
+ 3t
+x=0
3
2
t
t
t
...
which reduces to x + ẍ + ẋ + x = 0. The corresponding characteristic
equation is m3 + m2 + m + 1 = (m + 1)(m2 + 1) = 0. Therefore, the
general solution is
x = c1 e
s
+ c1 sin s + c2 cos s =
c1
+ c2 sin(ln t) + c3 cos(ln t),
t
t > 0.
34. Show that if x1 (t) satisfies the equation x000 +p1 (t)x00 +p2 (t)x0 +p3 (t)x =
0, then the substitution x = vx1 reduces the order of the equation from
3 to 2.
Solution. Substituting x = vx1 , we obtain
00 0
0 00
000
00
0 0
00
0
0
(x000
1 v+3x1 v +3x1 v +x1 v )+p1 (x1 v+2x1 v +x1 v )+p2 (x1 v+x1 v )+p3 x1 v = 0.
By regrouping terms, we obtain
00
0
00 0
0 00
000
0 0
00
0
v(x000
1 +p1 x1 +p2 x1 +p3 x1 )+3x1 v +3x1 v +x1 v +p1 (2x1 v +x1 v )+p2 x1 v = 0.
The last equation is a second order equation in v 0 . Actually, if we let
w = v 0 , it can be written as
x1 w00 + (3x01 + p1 x1 )w0 + (3x001 + 2p1 x01 + p2 x1 )w = 0.
88
CHAPTER 6. SOLUTIONS TO EXERCISES OF CHAPTER 6
Chapter 7
Solutions to exercises of
Chapter 7
0
✓
1
0
0
3
◆
A1. Solve x = Jx where J =
✓ ◆
x
Solution. If x =
the system becomes
y
✓ 0 ◆ ✓
◆✓ ◆
x
1
0
x
=
, namely
0
0
3
y
y
Solving we find x = c1 et , y = c2 e
3t
⇢
x0 = x
y 0 = 3y
.
We can find the same result noticing that the eigenvalues of J are
= 1, 3 and hence the general solution is
✓ ◆
✓ ◆
✓ ◆ ✓
◆
1
0
x
c1 e t
t
3t
x = c1
e + c2
e , namely
=
,
0
1
y
c2 e 3t
as before.
0
✓
a 0
0 a
◆
A2. Solve x = Jx where J =
, with a 6= 0.
✓ ◆
x
Solution. If x =
the system becomes
y
✓ 0 ◆ ✓
◆✓ ◆
⇢ 0
x
a 0
x
x = ax
=
, namely
y0
0 a
y
y 0 = ay.
89
90
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Solving we find x = c1 eat , y = c2 eat .
We can find the same result noticing that the matrix J has a double
eigenvalues = a. Since J is diagonal the general solution is
✓ ◆
✓ ◆
✓ ◆ ✓
◆
1
0
x
c1 eat
at
at
x = c1
e + c2
e , namely
=
0
1
y
c2 eat
0
A3. Solve x = Jx where J =
✓
a 1
0 a
◆
, with a 6= 0.
Solution. The system has the form
✓ 0 ◆ ✓
◆✓ ◆
x
a 1
x
=
y0
0 a
y
The matrix J has a double eigenvalue
= a, but is not diagonal.
Corresponding to = a, one eigenvector is
✓ ◆
1
v=
0
and hence one solution is
x1 =
✓
eat
0
◆
To find another linearly independent solution, we let
✓ ◆
✓
◆
1
u1
at
at
x2 = te
+e
0
u2
and solve (A
I)ū = v, for the components u1 , u2 of ū. We see that
we can take u1 = 0 and u2 = 1. Therefore we obtain
✓ ◆
✓ ◆
1
0
at
at
x2 = te
+e
0
1
Thus the general solution is
✓
◆
c1 eat + c2 teat
x=
.
c2 eat
91
The same result can be found if we notice that the system can be
written as
⇢ 0
x = ax + y
y 0 = ay.
From the second equation we get y = c2 eat . Substituting in the first we
find x0 = ax + c2 eat that can be solved finding again x = c1 eat + c2 teat .
✓
◆
3
1
0
A4. Solve x̄ = J x̄, J =
.
1 3
Solution. The characteristic equation is (3
)2 + 1 = 0. Hence
3
= ±i and = 3 ± i. Corresponding to = 3 + i, we solve
✓
◆✓ ◆
✓ ◆
3
1
x
x
= (3 + i)
.
1 3
y
y
which is equivalent to the system
⇢
3x y = 3x + ix
x + 3y = 3y + iy
If we let x = 1 then y =
i so that the corresponding eigenvector is
✓ ◆
1
v¯1 =
.
i
Similarly, corresponding to
=3
v¯2 =
✓
i, we have
◆
i
.
1
So, recalling that e(↵+ i)t = e↵t (cos t + i sin t), corresponding to
3 + i, we have the solution
✓ ◆ ✓ 3t
◆
1
e cos t + ie3t sin t
3t
e (cos t + i sin t)
=
=
i
ie3t cos t + e3t sin t
✓ 3t
◆
✓ 3t
◆
e cos t
e sin t
=
+i
.
e3t sin t
e3t cos t
=
The real and imaginary parts of the above gives us two linearly independent solutions, and the general solution is then
✓ 3t
◆
✓ 3t
◆ ✓ 3t
◆
e cos t
e sin t
c1 e cos t + c2 e3t sin t
x̄ = c1 3t
+ c2
=
.
e sin t
e3t cos t
c1 e3t sin t c2 e3t cos t
92
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
A5. Solve
✓
x0
y0
◆
=J
✓
x
y
◆
,
J=
✓
2 1
1 2
◆
.
Solution. The eigenvalues of J are = 2 ± i. The eigenvector corresponding to 2 + i is obtained by solving
⇢
ai + b = 0
=) a = 1, b = i,
a bi = 0
which yields a solution as
✓
◆
✓ ◆ ✓ 2t
◆
x(t)
1
e (cos t + i sin t)
(2+i)t
=e
=
y(t)
i
ie2t (cos t + i sin t)
✓ 2t
◆ ✓
◆
✓ 2t
◆
e (cos t + i sin t)
e2t cos t
e sin t
=
=
+i
.
e2t (i cos t sin t)
e2t sin t
e2t cos t
Two real, linearly independent solutions are
✓
◆
✓ 2t
◆
e2t cos t
e sin t
x1 =
,
x2 =
.
e2t sin t
e2t cos t
and the general solution is x = c1 x1 + c2 x2 namely
x(t) = e2t (c1 cos t + c2 sin t)
y(t) = e2t ( c1 sin t + c2 cos t)
0
A6. Solve x = Jx, where J =
✓
1 3
3 1
◆
.
Solution. Here the eigenvalues of J are 1 ± 3i. Thus the general
solution is
✓ t
◆
e [c1 cos 3t + c2 sin 3t]
x=
.
et [ c1 sin 3t + c2 cos 3t]
A7. Solve the Cauchy problem
✓
◆
1 0
0
x =
x,
0 1
x(0) =
✓
1
1
Solution. The general solution of the system is
✓ ◆
✓ ◆
1
0
t
x = c1
e + c2
et .
0
1
◆
.
93
Imposing the initial conditions we find
✓ ◆
✓ ◆ ✓
◆
1
0
1
c1
+ c2
=
=) c1 = 1, c2 =
0
1
1
1.
Otherwise, we can notice that the system is equivalent to
⇢ 0
x = x, x(0) = 1,
y 0 = y,
y(0) = 1.
Solving we find x = e t , y =
et .
A8. Solve the Cauchy problem
✓
◆
0
2
0
x =
x,
2
0
x(0) =
✓
a
0
◆
.
Solution. The eigenvalues of A are ±2i, and thus the general solution
is
✓
◆
c1 cos 2t + c2 sin 2t
x=
.
c1 sin 2t c2 cos 2t
The initial conditions yield
✓
◆ ✓ ◆
c1
a
=
=) c1 = a, c2 = 0.
c2
0
We can also notice that the system is equivalent to
⇢ 0
x = 2y, x(0) = a,
y 0 = 2x,
y(0) = 0,
or x00 + 4x, x(0) = a, x0 (0) = 0 Solving we find x = a cos 2t, y = a sin 2t,
as before.
A9. Solve
⇢
x0 =
y0 =
y, x(0) = 0
x, y(0) = 1
0
Solution. The system has the form x = Jx where J =
The eigenvalues of J are ±i and the general solution is
✓
◆ ✓
◆
x(t)
c1 cos t + c2 sin t
x=
=
y(t)
c1 sin t + c2 cos t
✓
0 1
1 0
◆
.
94
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
From x(0) = 0, y(0) = 1 we infer
✓ ◆ ✓
◆
0
c1
=
=) c1 = 0, c2 = 1.
1
c2
Thus
✓
x(t)
y(t)
◆
=
✓
sin t
cos t
◆
.
The same result (actually, in a more direct way) can be found if we
notice that the given problem is equivalent to the Cauchy problem
x00 + x = 0, x(0) = 0, x0 (0) = 1, whose solution is x(t) = sin t, y(t) =
x0 (t) = cos t.
A10. Solve
⇢
x0 = 3x + t
y0 =
y + 2t
Solution. The general solution of the associated homogeneous system
is
⇢
x = c1 e3t
y = c2 e t
Looking for a particular solution in the form xp = at + b, yp = ct + d,
we get
⇢
⇢
a=
3(at + b) + t
a = b + (3a + 1)t
=)
c=
(ct + d) + 2t
c=
d + (2 c)t
The first equation implies that a = b and
yields c = d and c = 2. Thus
(
1
x = c1 e3t
t
3
y = c2 e t + 2t
A11. Solve
⇢
a =
1
while the second
3
1
3
2
x0 = x + t 2
y0 = y + 1
Solution. The general solution of the associated homogeneous system
is
⇢
x = c1 e t
y = c2 e t
95
Looking for✓a particular solution
of the non homogeneous system of the
◆
2
at + bt + c
form xp =
, a, b, c, d 2 R, we find
d
⇢
2at + b = at2 + bt + c + t2
0 =d+1
Rearranging,
⇢
(a + 1)t2 + (b
2a)t + c
b =0
d = 1
Solving the first algebraic equation we get a + 1 = 0, 2a = b, b = c.
Thus
⇢
x = c1 et t2 2t 2
y = c2 e t 1
A12. Solve
⇢
x0 =
y0 =
x+y+1
x+y 5
Solution. The general solution of the associated homogeneous system
is
✓ t
◆
e [c1 cos t + c2 sin t]
x=
.
et [ c1 sin t + c2 cos t]
Looking for✓a particular
solution of the non homogeneous system of the
◆
a
form xp =
, a, b 2 R, we find
b
⇢
0=
0=
a+b+1
a+b 5
Solving this algebraic system we have b = 2, a = 3. Thus
⇢
x=
et [c1 cos t + c2 sin t] 3
y = et [ c1 sin t + c2 cos t] + 2
B1. Show that the matrix
A=
✓
a b
c d
◆
96
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
has 0 as an eigenvalue if and only if A is singular.
Solution. Setting the characteristic polynomial equal to zero, we have
✓
◆
a
b
= ad (a + d)
bc + 2 = 0
c
d
It is clear that if = 0 is a root, then ad
then = 0 is a root.
bc = 0 and if ad
bc = 0
B2. Show that the symmetric matrix
A=
✓
a b
b c
◆
has has two distinct real eigenvalues if b 6= 0.
Solution. The characteristic equation is (a
)(c
) b2 = 0 which
can be written as 2 (a + c) + ac b2 = 0. The discriminant of
this quadratic equation is (a + c)2 4(ac b2 ) = a2 2ac + c2 + 4b2 =
(a c)2 + 4b2 > 0.
B3. Find the eigenvalues and the corresponding eigenvectors of the matrix
✓
◆
1 0
A=
3 1
and write the general solution of the system x̄0 = Ax̄.
Solution. A is triangular and
✓ ◆its eigenvalues are
a
to find an eigenvector v1 =
corresponding to
b
✓
◆✓ ◆ ✓ ◆
2 0
a
0
=
3 0
b
0
= 1,
1. In order
= 1, we solve
which is equivalent to the system
⇢
2a = 0
3a = 0.
We see that a = 0 and b can be chosen arbitrarily. We choose, the
eigenvector
✓ ◆
0
v1 =
.
1
97
Corresponding to = 1, we have only one equation, 3a + 2b = 0 to
solve. Let us choose a = 2. Then b = 3 and hence
✓ ◆
2
v2 =
3
The general solution of the di↵erential equation is
✓ ◆
✓
◆ ✓
◆
0
2e t
2c2 e t
x̄(t) = c1 t + c2
=
3e t
c1 et 3c2 e t
e
B4. Find the general solution of the system
⇢ 0
x = 2x + 6y
y 0 = x + 3y.
Solution. Writing the given system
✓ 0◆ ✓
x
2
=
0
1
y
in matrix form, we have
◆✓ ◆
6
x
.
3
y
The eigenvalues of this matrix are = 0, 5. Corresponding to
we take the eigenvector
✓ ◆
3
v1 =
1
and corresponding to
= 5, we take
✓ ◆
2
v2 =
.
1
The general solution is
✓ ◆ ✓
◆
x
3c1 + 2c2 e5t
=
.
y
c1 + c2 e5t
B5. Find the general solution of
⇢ 0
x = 2x + 6y + et
y 0 = x + 3y et
= 0,
98
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Solution. From the previous exercise we know that the general solution of the homogeneous system is
✓ ◆ ✓
◆
x
3c1 + 2c2 e5t
=
.
y
c1 + c2 e5t
Let us find a particular solution setting xp = aet , yp = bet . We find
⇢ t
ae = 2aet + 6bet + et
bet = aet + 3bet et
and hence
⇢
that is b =
a = 2a + 6b + 1
b = a + 3b 1
namely
⇢
a + 6b =
a + 2b =
1
1
1
, a = 2. Thus
2
x(t) = 3c1 + 2c2 e5t + 2et ,
y(t) =
c1 + c2 e5t
1 t
e.
2
B6. Solve the initial value problem
✓ 0◆ ✓
◆✓ ◆ ✓
◆ ✓ ◆
x
1 2
x
x(0)
2
=
,
=
.
y0
0 3
y
y(0)
3
Solution. Calculating the eigenvalues, we find 1 = 1 and
Corresponding to 1 = 1, we can take the eigenvector to be
✓ ◆
1
v1 =
0
and corresponding to
2
= 3, we take the eigenvector
✓ ◆
1
.
1
Therefore, the general solution is
✓ t
◆
c1 e + c2 e3t
x̄ =
.
c2 e3t
2
= 3.
99
Now, in order to find the desired solution, we let t = 0 and solve for c1
and c2 in the equation
✓
◆ ✓ ◆
c1 + c2
2
=
c2
3
obtaining c1 =
1 and c2 = 3. So, the solution is
✓ t
◆
e + 3e3t
x̄ =
.
3e3t
B7. Solve
⇢
x0 = x + 2y + 2t
y 0 = 3y + t2
Solution. The general solution of the homogeneous system is
✓ ◆
✓ ◆
✓ ◆
x
1
1
t
= c1
e + c2
e3t
y
0
1
We try to find a particular solution in the form xp = a1 + a2 t + a3 t2 ,
yp = b1 + b2 t + b3 t2 . One finds
a2 + 2a3 t = a1 + a2 t + a3 t2 + 2(b1 + b2 t + b3 t2 ) + 2t
b2 + 2b3 t = 3(b1 + b2 t + b3 t2 ) + t2
namely
a2
2b1 = (a2 2a3 + 2b2 + 2)t + (a3 + 2b3 )t2
3b1 = (3b2 2b3 )t + (3b3 + 1)t2
a1
b2
Using the principle of identity of polynomials we find
a2
a1
b2
2b1 = 0, a2 2a3 + 2b2 + 2 = 0, a3 + 2b3 = 0,
3b1 = 0, 3b2 2b3 = 0,
3b3 + 1 = 0.
Solving, we get
b3 =
1
,
3
2
b2 = b 3 =
3
2
,
9
1
b1 = b 2 =
3
2
27
100
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
and
2
2b3 = ,
3
a3 =
a2 = 2a3
2b2
a1 = a2
2b1 =
4 4
+
3 9
2
4
+
=
9 27
2=
2=
2
.
27
2
,
9
In conclusion,
x(t) = c1 et + c2 e3t
2
27
y(t) = c2 e3t
B8. Solve
✓
x0
y0
◆
=A
✓
x
y
◆
2
27
2
t
9
,
2
2
t + t2 ,
9
3
1 2
t.
3
A=
✓
3
0
1
2
◆
.
Solution. The eigenvalues of the coefficient matrix A are = 2, 3.
Corresponding
to ✓ = 2,
= 3, we take the eigenvector v 1 =
✓ ◆
◆ resp.
1
1
, resp. v 2 =
. The general solution is
1
0
✓ ◆
✓ ◆
✓ ◆
x
1
1
2t
= c1
e + c2
e3t .
y
1
0
B9. Solve
✓
x0
y0
◆
=A
✓
◆
x
y
✓
1
4
A=
✓
where A =
0
1
◆
.
= ±1.✓Corresponding
to✓ = ◆1,
◆
1
0
resp. = 1, we take the eigenvector v 1 =
, resp. v 2 =
.
2
1
The general solution is
✓ ◆
✓ ◆
1
0
t
x(t) = c1
e + c2
e t.
2
1
Solution. The eigenvalues of A are
B10. Solve
0
x = Ax,
1 0
3 1
◆
101
Solution. The matrix
= 1. A first
✓ ◆ A has a double eigenvalue
a
eigenvector v 1 =
is obtained by solving (A I)v 1 = 0 which
b
✓ ◆
0
yields a = 0, b arbitrary, so that we can take v 1 =
yielding a
1
✓ ◆
0
first solution x1 =
et . A second linearly independent solution
1
may be found by substituting tet v̄1 +et v̄2 into the equation x0 = Ax and
solving for v2 (v̄1 is already found above), thus obtaining the second
solution
✓ ◆
✓ 1 ◆
0
t
x2 =
te + 3 et .
1
0
Thus the general solution is
✓
◆
✓ ◆
✓ ◆
✓ 1 ◆
x(t)
0
0
t
t
= c1
e + c2
te + c2 3 et
y(t)
1
1
0
One can check that the same result holds if we solve the equivalent
second order equation y 00 = 3x0 + y 0 = 3y + y 0 , or else if we solve the
first equation x0 = x yielding x = cet and substitute into y 0 = 3x + y.
B11. Solve
⇢
x0 = x + 3y,
y 0 = x y.
✓
◆
1
3
Solution. The eigenvalues of the matrix A =
are the roots
1
1
of (1
)( 1
) 3 = 0 namely 2 4 = 0, = ±2. Up to constants,
we can take the eigenvalues
✓ ◆
✓
◆
3
1
v̄1 =
, v̄2 =
.
1
1
Thus the general solution is
✓ ◆ ✓
◆
x
3c1 e2t + c2 e 2t
=
.
y
c1 e2t c2 e 2t
B12. Solve the Cauchy problem
⇢ 0
x = x + y, x(0) = 1
y 0 = x y, y(0) = 0
102
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
✓
0
x
y
◆
Solution. We first solve the system x = Ax, where x =
and
✓
◆
p
1
1
A=
. The eigenvalues of A are = ± 2. Corresponding
1
1
p
to = 2, we find the
a and b by solving
peigenvector with components
p
the equations
2) + b = 0, a b(1 + 2) = 0. Substituting
p a(1
b = a(1
2) into the second equation results in 2a
p 2a = 0. Hence
a can be chosen arbitrarily. Let a !
= 1. Then b = 2 1 and hence
p
2t
e
p
one solution is x̄1 = p
.
( 2 1)e 2 t
p
Similarly,
corresponding top =
2, we solve the equations (1 +
p
2)a + b = 0, a + b( 1 + 2) = 0. Here again, we obtain a a = 0.
So, a is again arbitrary. But in order to get a linearly independent
solution, this time we let a = 1, for
! example. Then we find the
p
2t
e
p
p
second solution x̄2 =
.
(1 + 2)e 2 t
B13. Solve the Cauchy problem
⇢ 0
x = x + y, x(0) = 1
y 0 = y,
y(0) = 2
Solution. Taking advantage from the fact that the second equation is
independent of x, we first solve the second problem yielding y = 2e t .
Substituting into the first equation we find x0 = x + 2e t and hence
x = cet e t . The initial condition x(0) = 1 implies c = 0. Thus
x = e t . The student could find the same result finding the general
solution of the system and imposing the initial conditions.
B14. Solve
⇢
x0 = x + 3y + 2t,
y 0 = x y + t2 .
Solution. From Ex. B11 we know that the solution of the associated
homogeneous system is
x = 3c1 e2t + c2 e
2t
,
y = c1 e2t
c2 e
2t
.
103
Le us find a particular solution in the form xp = a1 + a2 t + a3 t2 , yp =
b1 + b2 t + b3 t2 . One finds
a2 + 2a3 t = a1 + a2 t + a3 t2 + 3b1 + 3b2 t + 3b3 t2 + 2t,
b2 + 2b3 t = a1 + a2 t + a3 t2 b1 b2 t b3 t2 + t2 .
namely
a2 a1 3b1 = (a2 + 3b2 2a3 + 2)t + (a3 + 3b3 )t2 ,
b2 a1 + b1 = (a2 b2 2b3 )t + (a3 b3 + 1)t2 .
Then
a2
b2
a1 3b1 = 0, a2 + 3b2 2a3 + 2 = 0, a3 + 3b3 = 0,
a1 + b1 = 0, a2 b2 2b3 = 0, a3 b3 + 1 = 0.
Solving
7
, b1 =
8
Thus the particular
a1 =
xp =
B15. Solve
1
1
, a2 =
,
8
2
solution is
7 1
3 2
t
t,
8 2
4
⇢
b2 =
1. a3 =
yp =
1
8
t+
3
,
4
1
b3 = .
4
1 2
t.
4
x0 = x + 2y + et ,
y 0 = x 2y et .
p
Solution. The eigenvalues of the coefficient✓ matrix
are
=
±
2.
1,2
◆
ai
To find the corresponding eigenvectors vi =
we solve
bi
p
p
⇢
⇢
(1
2)a1 p
+ 2b1 = 0
(1 + 2)a2 p
+ 2b2 = 0
a1 2(1 + 2)b1 = 0
a2 2(1
2)b2 = 0
✓
◆
2
p
Up to constants, we can take the eigenvectors v1 =
and
2 1
✓
◆
2
v2 = p
. Then the solution of the homogeneous system is
2+1
p
p
p
p
p
p
x = 2c1 e 2 t 2c2 e 2 t , y = c1 ( 2 1)e 2 t + c2 ( 2 + 1)e 2 t .
We can seek a particular solution in the form xp = Aet yp = Bet ,
finding A = B = 12 and hence xp = 12 et and yp = 12 et .
104
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
C1. Show that if A and B are n ⇥ n matrices and B is invertible, then A
and B 1 AB have the same eigenvalues.
Solution. Suppose is an eigenvalue of A and let Ax̄ = x̄. Then
A(B 1 B)x̄ = x̄. Multiplying both sides on the left by B, we can write
the above equation as (BAB 1 )(B x̄) = B x̄. Let B x̄ = v̄. Then we
have (BAB 1 )v̄ = v̄. Therefore, is an eigenvalue of BAB 1 .
C2. Solve
0
1
0 1
x0
x
@ y0 A = A @ y A ,
z0
z
0
1
0 0
1 0 A.
0 2
1
1
2
A=@
Solution. The eigenvalues
0
1 of A are i = 1, 1, 2. The corresponding
ai
eigenvectors v i = @ bi A are the solutions of (A
i I)v i = 0, namely
ci
8
8
8
0 =0
2a2 = 0
a3 = 0
<
<
<
a1 2b1 = 0 ,
a2 = 0
a3 3b3 = 0
:
:
:
2a1 + c1 = 0
2a2 + 3c2 = 0
2a3 = 0
We find
0
1
2
v1 = @ 1 A ,
4
0
1
0
v2 = @ 1 A ,
0
1
0
v3 = @ 0 A
1
and hence the general solution is
0
1
0 1
0 1
x(t)
2
0
@ y(t) A = c1 @ 1 A et + c2 @ 1 A e
z(t)
4
0
0
1
1 0 0
C3. Solve x0 = Ax, where A = @ 0 2 1 A.
1 0 3
0
0
1
0
t
+ c3 @ 0 A e2t .
1
Solution. To find the eigenvalues of A we solve
1
0
0 2
1
0 3
0
1
= (1
)(2
)(3
) = 0.
105
Thus0the eigenvalues
are = 1, 2, 3. The corresponding eigenvalues
1
ai
v i = @ bi A (i = 1, 2, 3) solve
ci
0
10
1
⇢
0 0 0
a1
b1 + c 1 = 0
@ 0 1 1 A @ b1 A = 0 =)
a1 + 2c1 = 0
1 0 2
c1
8
0
10
1
1 0 0
a2
a2 = 0
<
@ 0 0 1 A @ b2 A = 0 =)
c2 = 0
:
1 0 1
c2
a2 + c 2 = 0
8
0
10
1
2
0 0
a3
2a3 = 0
<
@ 0
A
@
A
1 1
b3
b3 + c3 = 0
= 0 =)
:
1
0 0
c3
a3 = 0
Thus we can take
0
v̄1 = @
1
2
1 A,
1
and the general solution is
0
1
0
v̄2 = @ 1 A ,
0
0
1
0
v̄3 = @ 1 A
1
x = c1 v̄1 et + c2 v̄2 e2t + c3 v̄3 e3t .
C4. Find x solving the Cauchy
0
1
0
@
0
x =
0
problem
1
0 1
1 0 A x,
0 4
0
1
1
x(0) = @ 0 A
1
Solution. In terms of components the system is
8 0
< x1 = x1 + x3
x0 = x2
: 20
x3 = 4x3
The second and third equations can be solved separately, yielding x2 =
c2 e t , x3 = c3 e4t . Imposing the initial conditions we find c2 = 0, c3 = 1.
The first equation becomes x01 = x1 + e4t yielding x1 = 13 e4t + c1 et .
The initial condition x1 (0) = 1 implies that c1 = 23 . Therefore, x1 =
1 4t
e + 23 et . Thus the solution is x1 = 13 e4t + 23 et , x2 = 0, x3 = e4t
3
106
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
0
1
1 0 0
C5. Solve x0 = Ax, where A = @ 0 2 1 A.
1 0 1
Solution. In this case the
are = 1 (double) and = 2.
0 eigenvalues
1
a
A first eigenvector v 1 = @ b A related to = 1 is found by solving
c
We can take v 1
obtained solving
0
namely
@
⇢
0
b+c = 0
a = 0
1
0
= @ 1 A. A second
1
v 1 + v 2 = Av 2 , namely
1 0 0 1 0
0
a
1
1 A + @ b0 A = @ 0
1
c0
1
8
<
:
Then one has
0
1
a0
eigenvector v 2 = @ b0 A is
c0
10 0 1
0 0
a
2 1 A @ b0 A
0 1
c0
a0 =
a0
1 + b0 = 2b0 + c0
1 + c 0 = a0 + c 0
⇢
1 + c 0 = b0
1 = a0
0
1
1
Thus we can take v 2 = @ 1 A. It remains to find an eigenvector
0
0 00 1
a
v 3 = @ b00 A related to = 2. One finds (A 2I)v 3 = 0, namely
c00
8
<
:
a00
a00 = 0
c00 = 0
c00 = 0
0
1
0
=) v 3 = @ 1 A
0
107
In conclusion the general solution is
0
1
0
1
0
0
0
x(t) = c1 @ 1 A et + c2 @ 1 A tet + c2 @
1
1
1
0 1
1
0
t
A
@
1 e + c3
1 A e2t .
0
0
C6. Recall that the characteristic equation of the di↵erential equation
x000
2x00 + 3x0 + x = 0
is m3 2m2 + 3m + 1 = 0. Change the di↵erential equation to a system
and then show that its characteristic equation as a system remains the
same.
Solution. Letting x1 = x, x01 = x2 , x02 = x3 , the scalar di↵erential
equation can be written as the system
0 01 0
10 1
x1
0
1 0
x1
@x02 A = @ 0
A
@
0 1
x2 A .
0
x3
1
3 2
x3
The characteristic equation of this system is given by
1
0
1
0
1
=
3
+2
2
3 + 1 = 0.
3 2
If we multiply both sides of this equation by 1, we obtain the characteristic equation for the scalar version, with the only di↵erence being
that m is now called .
C7. Solve the Cauchy problem
8 0
x(0) = 0
< x =x+z
0
y = y + z y(0) = 1
: 0
z =y z
z(0) = 0
Solution. The coefficient matrix is
0
1
0
@
0
1
A=
0
1
1
1
1 A
1
108
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Its characteristic equation is
1
0
0
0
= (1
1
1
1
1
)[( 1
=
1
)( 1
)
1] = (1
)(
2
+2 )=0
Thus the0
eigenvalues
are i = 0, 1, 2. The corresponding eigenvectors
1
ai
are v i = @ bi A such that (A
i I)v i = 0̄ namely
ci
Av 1 = 0̄,
(A
I)v 2 = 0̄,
The first systems yields
8
< a1 + c 1 = 0
b1 + c 1 = 0
:
b1 c 1 = 0
Similarly, (A
I)v 2 = 0̄ yields
8
c2 = 0
<
2b2 + c2 = 0
:
b2 2c2 = 0
Finally, (A + 2I)v 3 = 0̄ yields
8
< 3a3 + c3 = 0
b3 + c 3 = 0
:
b3 + c 3 = 0
The general solution is
0 1
0
x
@ y A = c1 @
z
(A + 2I)v 3 = 0̄
0
=) v 1 = @
1
1
1 A
1
0
1
1
=) v 2 = @ 0 A .
0
0
=) v 3 = @
1
3
1
1 A.
1
1
0 1
0
1
1
1 A + c2 @ 0 A e t + c3 @
1
0
1
3
1
1 Ae
1
2t
Setting t = 0 and imposing the initial conditions, we find the system
8
8
< 0 = c1 + c2 + 13 c3
< c1 = 12
1 = c1 + c3
c2 = 13
=)
:
:
0 = c1 c3
c3 = 12
109
In conclusion, the solution of the Cauchy problem is
0 1
0
1
0 1
0 1 1
x
1
1
3
@ y A = 1 @ 1 A + 1 @ 0 A et + 1 @ 1 A e
2
3
2
z
1
0
1
2t
namely
x=
1 1 t 1
+ e + e
2 3
6
2t
,
y=
C8. Find a 2 R such that the system
x0 = Ax,
1 1
+ e
2 2
2t
,
z=
1
2
1
e
2
2t
.
0
1
a 0 0
A = @ b1 b2 0 A
b4 b5 b6
has a nontrivial solution x(t) satisfying |x(t)| ! 0 as t ! +1 for all
bi 2 R.
Solution. The eigenvalues of A are
= a, b2 , b6 . If v 6= 0 is an
eigenvector corresponding to a, the function x(t) = veat is a solution
of the system. If a < 0 we have that |x(t)| = |v|eat ! 0 as t ! +1.
0
1
2 0
0
1 A, find a non trivial solution of x0 = Ax such
C9. If A = @ 0 2
1 0
1
that limt!+1 |x(t)| = 0.
Solution. One eigenvalue of A is 1, and a corresponding solution is
x(t) = ve t , where (A + I)v = 0. This equation yields
8
3a = 0
<
3b + c = 0
:
a = 0
0
1
0
and we can take v = @ 1 A. Thus a solution which tends to zero at
0
1 3
0
@
1 A e t.
+1 is x(t) =
3
110
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
C10. Find a such that all the solutions of
0 0 1 0
x1
a 2
1
@ x02 A = @
1 a 2
x03
0
0
satisfy limt!+1 |xi (t)| = 0, i = 1, 2, 3.
Solution. The coefficient matrix is
0
1
0
1
A=@ 0
0
1
Its characteristic equation is
1
0
0
0
= (1
1
1
1
1
)[( 1
10
1
0
x1
0 A @ x2 A
a
x3
1
1
1 A
1
=
1
)( 1
)
1] = (1
)(
2
+2 )=0
Thus the0
eigenvalues
are i = 0, 1, 2. The corresponding eigenvectors
1
ai
are v i = @ bi A such that (A
i I)v i = 0̄ namely
ci
Av 1 = 0̄,
(A
I)v 2 = 0̄,
The first systems yields
8
< a1 + c 1 = 0
b1 + c 1 = 0
:
b1 c 1 = 0
Similarly, (A
I)v 2 = 0̄ yields
8
c2 = 0
<
2b2 + c2 = 0
:
b2 2c2 = 0
Finally, (A + 2I)v 3 = 0̄ yields
8
< 3a3 + c3 = 0
b3 + c 3 = 0
:
b3 + c 3 = 0
(A + 2I)v 3 = 0̄
0
=) v 1 = @
1
1
1 A
1
0
1
1
=) v 2 = @ 0 A .
0
0
=) v 3 = @
1
3
1
1 A.
1
111
The general solution is
0 1
0
x
@ y A = c1 @
z
1
0 1
0
1
1
1 A + c2 @ 0 A e t + c3 @
1
0
1
3
1
1 Ae
1
2t
Setting t = 0 and imposing the initial conditions, we find the system
8
8
< 0 = c1 + c2 + 13 c3
< c1 = 12
1 = c1 + c3
c2 = 13
=)
:
:
c3 = 12
0 = c1 c3
In conclusion, the solution of the Cauchy problem is
0 1
0
1
0 1
0 1 1
x
1
1
3
@ y A = 1 @ 1 A + 1 @ 0 A et + 1 @ 1 A e
2
3
2
z
1
0
1
2t
namely
x=
1 1 t 1
+ e + e
2 3
6
2t
,
y=
C8. Find a 2 R such that the system
x0 = Ax,
1 1
+ e
2 2
2t
,
z=
1
2
1
e
2
2t
.
0
1
a 0 0
A = @ b1 b2 0 A
b4 b5 b6
has a nontrivial solution x(t) satisfying |x(t)| ! 0 as t ! +1 for all
bi 2 R.
Solution. The eigenvalues of A are
= a, b2 , b6 . If v 6= 0 is an
eigenvector corresponding to a, the function x(t) = veat is a solution
of the system. If a < 0 we have that |x(t)| = |v|eat ! 0 as t ! +1.
0
1
2 0
0
1 A, find a non trivial solution of x0 = Ax such
C9. If A = @ 0 2
1 0
1
that limt!+1 |x(t)| = 0.
112
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Solution. One eigenvalue of A is 1, and a corresponding solution is
x(t) = ve t , where (A + I)v = 0. This equation yields
8
3a = 0
<
3b + c = 0
:
a = 0
0
1
0
and we can take v = @ 1 A. Thus a solution which tends to zero at
0
1 3
0
@
1 A e t.
+1 is x(t) =
3
C10. Find a such that all the solutions of
0 0 1 0
x1
a 2
1
@ x02 A = @
1 a 2
0
x3
0
0
10
1
0
x1
0 A @ x2 A
a
x3
satisfy limt!+1 |xi (t)| = 0, i = 1, 2, 3.
Solution. The system can be split into x03 = ax3 and
✓ 0 ◆ ✓
◆✓
◆
x1
a 2
1
x1
=
.
x02
1 a 2
x2
The eigenvalues of the last coefficient matrix are
the solutions
x1 = e(a
2)t
(c1 cos t + c2 sin t),
x2 = e(a
2)t
( c1 sin t + c2 cos t).
=a
2 ± i yielding
As for x3 we find x3 = c3 e at . The limit condition is satisfied provided
a > 0 and a 2 < 0, namely for 0 < a < 2.
D1. Solve
⇢
x0 + ty =
y 0 + x0 =
1,
2.
Solution. From the second equation we get y + x = 2t + c1 , and
hence y = x + 2t + c1 . Substituting into the first equation we find
113
x0 + t( x + 2t + c1 ) = 1, namely x0 tx = t2 c1 t
linear equation. Integrating we find

Z
1 2
1 2
t
x(t) = e 2 c2
e 2 t (t2 + c1 t 1)dt
1. This is a
and
y(t) =
x(t) + 2t + c1 = 2t + c1
D2. Solve
⇢
e
1 2
t
2

c2
Z
e
1 2
t
2
(t2 + c1 t
1)dt .
x0 + y = 3t,
y 0 tx0 = 0.
Solution. From the first eq. we get x0 = 3t y. Substituting into the
second, y 0 t(3t y) = 0, hence y 0 + ty = 3t2 . Integrating we find
✓
◆
Z
1 2
1 2
t
2
t
y=e 2
c1 + 3 t e 2 dt .
From x0 = 3t
y we infer x = 32 t2 +
D3. Solve
⇢
x0
y0
R
y(t)dt.
ty = 1,
tx0 = 3.
Solution. Multiplying the first equation by t we find tx0 = t2 y + t.
Then the second equation yields y 0 t2 y = t + 3. Solving, we find

Z
1 3
1 3
t
y(t) = e 3 c1 + e 3 t (t + 3)dt
and, from x0 = ty + 1,
x=
D4. Solve
⇢
Z
ty(t)dt + t + c2 .
t2 x0 y = 1,
y 0 2x = 0.
114
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Solution. From the first equation we find y = t2 x0 1. Substituting
into the second equation we get (t2 x0 )0 2x = 0, namely t2 x00 + 2tx0
2x = 0, a homogeneous Euler equation. Solving we find
y = t 2 x0
1 = c1 t 2
x0 y =
y 0 3x0 =
3,
2x.
x = c1 t + c2 t 2 ,
D5. Solve
⇢
1
2c2 t
1.
Solution. From the first equation we get y = x0 3. Substituting
into the second, x00 3x0 + 2x = 0, whose solution is x = c1 et + c2 e2t ,
whereby y = x0 3 = c1 et + 2c2 e2t 3.
D6. Solve
⇢
tx0 + y 0
= 1,
x0
y + x + e = 1.
0
Solution. Substituting y 0 = 1 tx0 in the second equation, we have:
0
0
1 tx0 + x + ex = 1, or tx0 x ex = 0. Di↵erentiating both sides,
we obtain
0
x00 (t ex ) = 0.
0
x00 = 0 leas to x0 = c1 . Substituting into tx0 x ex = 0 and solving one
has the solution x = c1 t ec1 . Moreover we get y 0 = 1 tx0 = 1 c1 t,
whereby y = t 12 c1 t2 + c2 . Thus
x = c1 t
e c1 ,
y=t
1 2
c1 t + c2
2
is one family of solutions depending on the constants c1 , c2 .
0
If x00 6= 0 then the problem reduces to solving t ex = 0. Taking
0
logarithm of both sides of the equation t = ex , we obtain the equations
x0 = ln t and y 0 = 1 tx0 = 1 t ln t. Integrating, we find for t > 0
x = t ln t t + c01 ,
y = t 12 t2 ln t + 14 t2 + c02 ,
which is another family of solutions depending on the constants c01 , c02 .
115
D7. Solve
⇢
xx0 + y 0 = 2t,
y 0 + 2x2 = 1.
Solution. Subtracting the second equation from the first we obtain
xx0 2x2 = 2t 1. Setting v = x2 , we get v 0 = 2xx0 = 2(2x2 + 2t 1) =
4v + 4t 2. Solving this linear equation we find v = c1 e4t t + 14 and
then
r
1
x = ± c1 e4t t + .
4
Furthermore,
y0 = 1
2x2 = 1
2v = 1
2c1 e4t + 2t
1
,
2
yielding
y=t
1 4t
c1 e + t 2
2
1
1
t + c2 = t
2
2
1 4t
c1 e + t 2 + c2 .
2
116
CHAPTER 7. SOLUTIONS TO EXERCISES OF CHAPTER 7
Chapter 8
Solutions to exercises of
Chapter 8
1. Find the equilibrium of
⇢
x0 = x + 1
y 0 = x + 3y
1
Solution. The equilibrium solves the algebraic system
⇢
0 =x+1
0 = x + 3y 1
One finds x =
2
1 and y = .
3
2. Find a, b such that the equilibrium of
⇢ 0
x = x + 3y + a
y0 = x y + b
is (1, 2).
Solution. x = 1, y = 2 is the equilibrium provided
⇢
0 =1+3·2+a
0 =1 2+b
whereby a =
7, b = 1.
117
118
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
3. Find ↵,
such that
is hamiltonian.
⇢
x0 = ↵x + y
y 0 = 2x + y
Solution. We need to see if there exists H(x, y) such that Hy = ↵x+y
and Hx = 2x+ y. Since Hyx = Hxy then ↵ =
. The hamiltonian
is H(x, y) = x2 + ↵xy + 12 y 2 .
4. Discuss the family of conics x2 + Bxy + y 2 = c in dependence on B, c.
Solution. The nature of the conics depends on the sign of B 2 4 and
c. Thus if B 2 < 4 and c > 0 the conics are ellipses, or circles if B = 0.
If B 2 > 4 and c 6= 0 the conics are hyperbola. If B = 2 or B = 2
we have (x + y)2 = c or (x y)2 = c, a pair of straight lines provided
c 0.
5. Discuss the family of conics Ax2
xy + y 2 = c in dependence on A, c.
Solution. If 1 4A < 0, that is if 4A > 1 and c > 0 we have an ellipse.
If 4A < 1 and c 6= 0 we have a hyperbola. If A = 14 , then the equation
is ( 12 x y)2 = c, a pair of parallel straight lines if c > 0 or the straight
line y = 12 x if c = 0.
6. Find C such that the system
⇢ 0
x =x+y
y 0 = 2Cx
y
has no periodic solution but the equilibrium x(t) = y(t) ⌘ 0.
Solution. The hamiltonian is H(x, y) = Cx2 + xy + 12 y 2 . If 1 2C > 0
the conic Cx2 + xy + 12 y 2 = c is not an ellipse. Thus if C < 12 the
solutions cannot be periodic.
7. Show that if AC < 0 then all the non trivial solutions of the system
⇢ 0
x = Bx + Cy
y 0 = Ax By
are not periodic.
Solution. The hamiltonian is H(x, y) = 12 Ax2 + Bxy + 12 Cy 2 . Since
B 2 AC > 0 the conic 12 Ax2 + Bxy + 12 Cy 2 = c is not an ellipse.
119
8. Find B such that the system
⇢ 0
x = Bx + 3y
y 0 = 3x By
has periodic solutions.
Solution. The conic 32 x2 + Bxy + 32 y 2 = c is an ellypse provided c > 0
and B 2 9 < 0.
9. Show that the solution of the system
8 0
x
=x+y
>
>
< 0
y
= 2x y
x(0)
=
1
>
>
:
y(0) = 0
is periodic.
Solution. The solutions verify H(x, y) = x2 + xy + 12 y 2 = c. For
x = 1, y = 0, one has c = 1. The curve x2 + xy + 12 y 2 = 1 is an ellipse
and corresponds to a periodic solution such that x(0) = 1, y(0) = 0.
10. Show that the solution of the system
8 0
x
= x 6y
>
>
< 0
y
= 2x y
x(0) = 1
>
>
:
y(0) = 0
is unbounded.
Solution. The solutions verify H(x, y) = x2 + xy 3y 2 = c. For x =
1, y = 0, one has c = 1. The curve x2 +xy 3y 2 = 1 is a hyperbola that
corresponds to an unbounded solution such that x(0) = 1, y(0) = 0.
11. Draw the phase plane portrait of the pendulum equation
Lx00 + g sin x = 0
and discuss the behavior of the solutions.
Solution. The conserved energy is
E=
L 2
y
2
g cos x,
120
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
y
c>1
c<1
c=1
O
-π
Figure 8.1: Plot of
L 2
y
2
π
x
g cos x = c
and curves of equation E = c are plotted in fig. 8.1.
There are infinitely many equilibria (xk , 0) where sin xk = 0, namely
xk = k⇡, k 2 Z. For c < 1 the solutions corresponding to the red curves
are periodic. If c = 1 the curve E = 1 are the black lines. They carry
heteroclinic solutions joining ⇡ and ⇡, or ⇡ and 3⇡, and so on. If
c > 1 the curves E = c are the blue lines. The solutions carried by the
branch contained in the half plane y > 0, resp. y < 0, are increasing,
resp. decreasing, and unbounded.
12. Find the equilibria of the Lotka-Volterra system
⇢ 0
x = x xy
y 0 = y + xy
Solution. The equilibria are the solutions of the algebraic system
⇢
0 = x xy
0 = y + xy
121
which are x = y = 0 and x = y = 1.
13. Find the nontrivial equilibrium (x✏ , y✏ ) of
⇢
x0 = 2x 7xy
y 0 = y + 4xy
✏x
✏y
Solution. One finds
x✏ =
1+✏
,
4
y✏ =
2
✏
7
.
14. Prove that there exists a periodic solution of the system
⇢
such that x + 2y
x0 = 2x 2xy
y 0 = y + xy
4 = ln(xy 2 ).
Solution. This is a Lotka-Volterra system with a = 2, b = 2, c = 1, d =
1. The solutions verify H(x, y) = x + 2y ln x 2 ln y = k, x, y > 0.
The equilibrium is (1, 1) and H(1, 1) = 3. For k > 3 the curve H = k is
(not empty and) compact, namely closed and bounded. In particular,
for k = 4 one has x + 2y = ln x + 2 ln y + 4, namely x + 2y 4 = ln(xy 2 ),
x, y > 0.
15. Prove that there exists a periodic solution of the system
⇢
such that x + 4y
x0 = x 4xy
y 0 = 2y + xy
4 = ln(x2 y).
Solution. This is a Lotka-Volterra system with a = 1, b = 4, c =
2, d = 1. The solutions verify H(x, y) = x + 4y 2 ln x ln y = k,
x, y > 0. The equilibrium is (2, 14 ) and H(2, 14 ) = 2 + 1 2 ln 2 ln 14 =
3 2 ln 2+ln 4 = 3+ln 2. For k > 3+ln 2 the curve H = k is (not empty
and) compact. In particular, for k = 4 one has x+4y 2 ln x ln y = 4,
namely x + 4y 4 = ln(x2 y).
122
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
16. Let x(t), y(t) be a T-periodic solution of the Lotka-Volterra system
⇢ 0
x = x(3 y)
y 0 = y(x 5)
RT
RT
Show that T1 0 x(t)dt = 5 and T1 0 y(t)dt = 3.
Solution. The nontrivial equilibrium of the system is x = 5, y = 3.
By the theory it follows that the mean value of x(t) is 5 and the mean
RT
RT
value of y(t) is 3, namely T1 0 x(t)dt = 5 and T1 0 y(t)dt = 3.
p
17. Show that U (t) = 2/ cosh t is a homoclinic of the equation x00 = x x3 .
Solution. By a direct calculation one finds that U 00 = U
over, one has that
et + e
cosh t =
2
U 3 . More-
t
! +1,
as t ! ±1
and hence U (t) ! 0 as t ! ±1.
18. Let x0 (t) be a homoclinic of x00 = x x3 . Show that x000
0 (t) ! 0 as
t ! ±1. Extend the result to any derivative of x0 (t).
Solution. Di↵erentiating the equation it follows that x000 = x0 3x2 x0 .
Since x0 (t) is a homoclinic then x0 (t) ! 0 and x00 (t) ! 0 as t ! ±1.
Hence also x000
0 (t) ! 0. Same argument for the higher order derivatives.
19. Prove the preceding result for the heteroclinincs of x00 =
Solution. As before, the result follows from x000 =
20. Show that the solution of x00 =
x + x3 .
x0 + 3x2 x0 .
x + x3 , x(0) = 0, x0 (0) =
1
2
2
is periodic.
1 4
Solution. The solutions of x00 = x + x3 satisfy y 2 + x
x = c.
2
Setting x = 0, y = 12 we find c = 14 . The corresponding curve passing
through (0, 12 ) is the inner part of the blue line plotted in fig. 8.2, which
is compact and does not contain any zero of f (x) = x + x3 . Hence
the solution is periodic.
21. Discuss the behavior of the solution of x00 =
2, x0 (0) = 0.
x + x3 such that x(0) =
Solution. Setting x = 2, y = 0 into y 2 + x2 12 x4 = c, we find c = 4.
The curve passing through (2, 0) is the branch of the red line in fig.
123
y
1
1/2
2
x
Figure 8.2: Plot of y 2 + x2
black.
1 4
x
2
= c, c =
1
4
in blue, c =
4 in red, c = 1 in
8.2 contained in the half plane x > 0. The corresponding solution
is positive (because x(t) > 0), increasing for t > 0 (because x0 (t) =
y(t) > 0 for t > 0) and decreasing for t < 0 (because x0 (t) = y(t) < 0
for t < 0). Hence x(t) has a minimum at t = 0, with x0 (0) = 0 and
x00 (0) = x(0) + x3 (0) = 6 > 0. Moreover limt!±1 x(t) = +1.
22. Discuss the behavior of the solutions x00 =
0, x0 (0) = 1.
x + x3 such that x(0) =
Solution. Setting x = 0, y = 1 into y 2 + x2 12 x4 = c, we find c = 1.
The curve y 2 +x2 12 x4 = 1 passing through (0, 1) is the part of the black
line in fig. 8.2 contained in the half plane y > 0. Since x0 (t) = y(t) > 0
for all t, the solution is increasing. Moreover, limt! 1 x(t) = 1 and
limt!+1 x(t) = +1.
23. Show that the solution of x00 = x
periodic.
x3 such that x(0) = 1, x0 (0) = 1 is
Solution. The conservation of the energy yields y 2
x2 + 12 x4 = c.
124
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
For x = 0, y = 1 we get c = 1. The curve y 2 x2 + 12 x4 = 1 passing
through (0, 1) is the red line in fig. 8.3 and is compact and does not
contain any zero of f (x) = x x3 . Thus the solution is periodic.
y
1
1/√2
Figure 8.3: Plot of y 2
black.
2
x
x2 + 12 x4 = c, c = 1 in red, c = 4 in blue, c =
24. Show that the solution of x00 = x
periodic.
3
8
in
x3 such that x(0) = 2, x0 (0) = 0 is
Solution. Putting x = 2, y = 0 in y 2 x2 + 12 x4 = c we get c = 4.
The curve y 2 x2 + 12 x4 = 4 passing through (2, 0) is the blue line in
fig. 8.3 and is compact and does not contain any zero of f (x) = x x3 .
Thus the solution is periodic.
p
25. Show that the solution of x00 = x x3 such that x(0) = 1/ 2, x0 (0) = 0
is periodic.
p
Solution. Putting x = 1/ 2, y = 0 in y 2 x2 + 12 x4 = c we get
c = 12 + 18 = 38 . The curve y 2 x2 + 12 x4 = 38 passing through
p
(1/ 2, 0) is the branch of the black line in fig. 8.3 contained in the half
125
plane x > 0. Such a curve is compact and does not contain any zero of
f (x) = x x3 . Thus the solution is periodic.
26. Show that for all a 6= 0 the solution of x00 + x + 8x7 = 0, x(0) = 0,
x0 (0) = a is periodic.
Solution. The conservation of the energy yields y 2 + x2 + 2x8 = c. For
x = 0, y = a we get c = a2 . Solving we find
p
y = ± a2 x2 2x8 .
Thus if a 6= 0 the curve y 2 + x2 + x8 = a2 is compact and does not
contain any zero of f (x) = x + 8x7 . Hence the solution is periodic.
27. Discuss the behavior of the solution of x00 + x + 13 x2 = 0, x(0) = 1,
x0 (0) = 0.
Solution. The conservation of the energy yieldspy 2 + x2 + 2x3 = c.
For x = 1, y = 0 we get c = 3 and hence y = ± 3 x2 2x3 . This
function is defined for 3 x2 2x3 0, namely for x  1 and is plotted
in fig. 8.4. It follows that the solution x(t) satisfies x(t)  1 for all t, is
increasing for t < 0 and decreasing for t > 0, and limt!±1 x(t) = 1.
Moreover, x(t) has its maximum at t = 0 with x(0) = 1, x0 (0) = 0 and
x00 (0) = x(0) 13 x2 (0) = 1 13 = 43 .
28. Show that the solution of x00
homoclinic to x = 0.
x + 3x2 = 0, x(0) = 12 , x0 (0) = 0 is
Solution. The conservation of the energy yields y 2 x2 + 2x3 = c.
The curve y 2 x2 + 2x3 = c is plotted in fig. 8.5. For x = 12 , y = 0 we
get c = 0. The curve y 2 x2 + 2x3 = 0 is the black line in fig. 8.5. The
branch contained in the half plane x > 0 carries a positive solution
x(t) with the following properties: (i) x(t) is even, (ii) x0 (t) > 0 for
t < 0 and x0 (t) < 0 for t > 0, (iii) limt!±1 x(t) = 0. This last property
shows that x(t) is homoclinic to 0.
29. Show that the solution of x00
periodic.
x + 3x2 = 0, x(0) = 14 , x0 (0) = 0 is
1
2
1
Solution. Now from y 2 x2 + 2x3 = c it follows c = 16
+ 64
= 32
.
The corresponding curve is the red line, plotted in Fig. 8.5. The
branch passing through ( 14 , 0) is compact and does not contain any
zero of f (x) = x 3x2 . Hence it gives rise to a periodic solution.
126
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
y
1
x
Figure 8.4: Plot of y 2 + x2 + 2x3 = 3.
30. Discuss the behavior of the solution of x00
x0 (0) = 1.
x + 3x2 = 0, x(0) = 0,
Solution. In this case c = 1. The curve y 2 x2 + 2x3 = 1 is the
blue curve in fig. 8.5 and carries the solution x(t) of the problem. Let
t0 > 0 be such that x(t0 ) = 1 and y(t0 ) = 0. Then x0 (t) > 0 for t < t0 ,
x0 (t0 ) = 0 and x0 (t) < 0 for t > t0 . Hence x(t) is increasing for t < t0 ,
decreasing for t > t0 and has its (unique) maximum at t0 . Moreover,
limt!±1 x(t) = 1.
127
y
1
1/4
1/2
x
1
Figure 8.5: Plot of y 2
x2 + 2x3 = c
128
CHAPTER 8. SOLUTIONS TO EXERCISES OF CHAPTER 8
Chapter 9
Solutions to exercises of
Chapter 9
1. If ↵ > 0, find the eigenvalues of x00 + ↵x = 0, x(0) = x(b) = 0.
Solution. From the theory we know that the eigenvalues have
p to be
positive,
namely
>
0.
The
general
solution
is
x(t)
=
c
sin
↵t +
1
p
↵ t. The condition
x(0) = 0 implies c2 = 0. The condition
c2 cos
p
x(b) = 0 yields c1 p
sin
↵ b = 0. To have
p a nontrivial solution we take
c1 6= 0. Then sin
↵ b = 0 implies
↵ b = k⇡, k a positive integer.
k2⇡2
Thus the eigenvalues are k =
, k = 1, 2, ....
↵b2
2. If is a positive number, find the eigenvalues of x00 + x = 0, x(0) =
x(b) = 0.
k2⇡2
Solution. Dividing by we find x00 + x = 0. Therefore k =
,
b2
k = 1, 2, ...
3. Estimate the eigenvalues of x00 + (1 + t)x = 0, x(0) = x(1) = 0.
Solution. Since 1  1 + t  2 in [0, 1], then the comparison property
of the eigenvalues yields k [2]  k [1 + t]  k [1]. Since
k [2] =
it follows
k2⇡2
,
2
k2⇡2

2
k [1
129
k [1]
= k2⇡2,
+ t]  k 2 ⇡ 2 .
130
CHAPTER 9. SOLUTIONS TO EXERCISES OF CHAPTER 9
4. Estimate the first eigenvalue of x00 + et x = 0, x(0) = x(2) = 0.
Solution. From 1  et  e2 for t 2 [0, 2], it follows that
⇡2
⇡2
t
2
and 1 [e ] = 2 then
1 [e ]  1 [1]. Since 1 [1] =
4
4e
⇡2

4e2
5. Show that the first eigenvalue
is smaller or equal to ⇡ 2 .
t
k [e ] 
1
1 [e
2
] 
⇡2
.
4
of (t2 x0 )0 + x = 0, x(0) = x(⇡) = 0
Solution. The variational characterization of the first eigenvalue yields
R ⇡ 2 02
t
(t)dt
0
1  R⇡ 2
(t)dt
0
2 C. Taking (t) = sin t one finds
R⇡ 2
t cos2 (t)dt
0R
.
1 
⇡
sin2 (t)dt
0
for all test function
Since 0  t  ⇡ it follows
1
because
R⇡
0
cos2 (t)dt =
R⇡
cos2 (t)dt
 ⇡ 2 R0⇡ 2
= ⇡2,
sin
(t)dt
0
R⇡
0
sin2 (t)dt.
6. If 0 < ↵  p(t) 
in [a, b], estimate the first eigenvalue
0 0
(p(t)x ) + x = 0, x(a) = x(b) = 0.
Solution. From
1
we infer
R⇡
↵
min R0 ⇡
2C
0
02
R⇡
p(t)
= min 0 R ⇡
2C
(t)dt

2 (t)dt
0
1
02
1
of
(t)dt
2 (t)dt
R⇡
 min R0 ⇡
2C
0
02
(t)dt
2 (t)dt
The above minima are, respectively, the first eigenvalues of ↵x00 + x =
0, x(a) = x(b) = 0, respectively x00 + x = 0, x(a) = x(b) = 0, which
are
↵⇡ 2
⇡2
,
[
]
=
.
1 [↵] =
1
(b a)2
(b a)2
131
Thus
↵⇡ 2

(b a)2
1

⇡2
.
(b a)2
7. Estimate the first eigenvalue 1 of (p(t)x0 )0 + q(t)x = 0, x(a) = x(b) =
0, under the assumption that 0 < ↵  p(t)  and 0 < m  q(t)  M
in [a, b].
Solution. Using the variational characterization of 1 , we find
R ⇡ 02
R ⇡ 02
↵ (t)dt
(t)dt
0
R0⇡
min R ⇡


min
1
2C
2C
M 2 (t)dt
m 2 (t)dt
0
0
whereby
↵⇡ 2

M (b a)2
1

⇡2
a)2
m(b
.
8. Let 1 [q], resp. e1 [q], be the first eigenvalue of (p(t)x0 )0 + q(t)x = 0,
resp. (e
p(t)x0 )0 + q(t)x = 0, with the boundary conditions x(a) =
x(b) = 0. If p(t)  pe(t) for all t 2 [a, b], show that 1 [q]  e1 [q].
Solution. The assumption p(t)  pe(t) in [a, b] implies
R⇡
R⇡
p(t) 02 (t)dt
pe(t) 02 (t)dt
0
R⇡
R0⇡

, 8 2 C.
q(t) 2 (t)dt
q(t) 2 (t)dt
0
0
Taking the minimum for
2 C, it follows that
1 [q]
 e1 [q].
9. Show that the eigenvalues of x00 + x = 0, x0 (a) = x0 (b) = 0 cannot be
strictly negative.
Solution.
If < 0 the general solution is x(t) = c1 e ↵t + c2 e↵t , where
p
↵=
. Then x0 = c1 ↵e ↵ t + c2 ↵e↵ t and x0 (a) = x0 (b) = 0 yield
⇢
c1 ↵e ↵ a + c2 ↵e↵ a = 0
c1 ↵e ↵ b + c2 ↵e↵ b = 0
The determinant of this system is
↵e
↵e
=
↵a
↵b
↵2 e
↵e↵ a
↵e↵ b
↵ (a b)
=
e
↵2 e
↵ (b a)
↵a ↵b
e
6= 0,
+ ↵2 e
↵b ↵a
e
132
CHAPTER 9. SOLUTIONS TO EXERCISES OF CHAPTER 9
because e ↵ (a b) 6= e ↵ (b a) whenever a 6= b. Therefore, the only solution of the system is c1 = c2 = 0 which corresponds to the trivial
solution.
10. Find the eigenvalues of x00 + x = 0, x0 (0) = x0 (⇡) = 0.
Solution. From the previous exercise
we knowpthat
0. Then the
p
general solution
is
x(t)
=
c
sin
t
+
c
cos
t,
0. One has
1 p
2
p
p
0
x (t) =
(c1pcos
t c2 sin
t).
p
p The boundary
p
pconditions yield
the system c1
= 0 and c1
cos
⇡ p
c2
sin
⇡ = 0, which
has a nontrivial solution provided = 0 or
= k, k = 1, 2, .... Then
2
the eigenvalues are k = k , with k = 0, 1, 2...
11. Find the eigenvalues of x00 + x = 0, x(0) = x0 (⇡) = 0.
Solution.
First, p
let
0. Then the general solution is x(t) =
p
c1 sin
t + c2 cos
t and the
= 0 yields
p boundary condition x(0)
0
c2 = 0, whereby x(t) = c1 sin
t,
0. The condition x (⇡) = 0 gives
p
p
p
⇡
⇡(2k + 1)
c1
cos
⇡ = 0,
0. Then = 0 or
⇡ = + k⇡ =
,
2
2
p
2k + 1
namely
=
, k = 0, 1, 2, .... The value = 0 yields x(t) ⌘ 0
2
and hence is not an eigenvalue.
p
If < 0, the general solution is x(t) = c1 e ↵t + c2 e↵t , where ↵ =
.
The condition x(0) = 0 yields c1 + c2 = 0. The condition x0 (⇡) = 0
yields ↵c1 e ↵⇡ + ↵c2 e↵⇡ = 0. The system
⇢
c1 + c2
=0
↵⇡
↵⇡
↵c1 e
+ ↵c2 e
=0
has only the trivial solution because its determinant is
1
↵e
↵⇡
1
↵e↵⇡
In conclusion, the eigenvalues are
= ↵e↵⇡ + ↵ 6= 0.
k
=
✓
2k + 1
2
◆2
, k = 0, 1, 2, ...
12. Let x(t) be a solution of the non-homogeneous problem x00 + k q(t)x =
h(t), x(a) = x(b) = 0, where k = k [q] is the k-th eigenvalue with
Rb
corresponding eigenfunction 'k . Prove that a h(t)'k (t)dt = 0.
133
Solution. Multiplying the equation by 'k and integrating we find
Z
b
00
x (t)'k (t)dt +
a
k
Z
b
q(t)x(t)'k (t)dt =
a
Z
b
h(t)'k (t)dt.
a
Integrating twice by parts and taking into account that x(a) = x(b) =
'k (a) = 'k (b) = 0 it follows
Z
b
00
x (t)'k (t)dt =
a
Z
b
a
x(t)'00k (t)dt.
Rb
Rb
00
Since '00k =
k q(t)'k we infer a x(t)'k (t)dt =
k a q(t)'k (t)dt.
Then
Z b
Z b
Z b
00
h(t)'k (t)dt =
x (t)'k (t)dt + k
q(t)x(t)'k (t)dt
a
a
a
Z b
Z b
00
=
x(t)'k (t)dt + k
q(t)x(t)'k (t)dt
a
a
Z b
Z b
=
q(t)x(t)'k (t)dt + k
q(t)x(t)'k (t)dt = 0.
k
a
a
13. Setting L(u) = (p(t)u0 )0 + r(t)u, show that L(u)v L(v)u = (p(uv 0
Rb
vu0 ))0 . Deduce that if u(a) = v(a) = u(b) = v(b) = 0, then a L(u)vdt =
Rb
L(v)udt.
a
Solution. One has that L(u)v = (pu0 )0 v + ruv and L(v)u = (pv 0 )0 u +
ruv. Subtracting, one finds L(u)v L(v)u = (pu0 )0 v (pv 0 )0 u = p0 u0 v +
pu00 v (p0 v 0 u + pv 00 u)=p0 (uv 0 vu0 ) + p(uv 00 vu00 )=(p(uv 0 vu0 ))0 .
Integrating, one finds
Z
b
(L(u)v
L(v)u)dt =
a
= p(b)(u(b)v 0 (b)
v(b)u0 (b))
Z
b
(p(uv 0
vu0 ))0 dt
a
p(a)(u(a)v 0 (a) v(a)u0 (a)).
Rb
Since u(a) = v(a) = u(b) = v(b) = 0 we deduce a [L(u)v L(v)u]dt =
Rb
Rb
0, whereby a L(u)vdt = a L(v)udt.
134
CHAPTER 9. SOLUTIONS TO EXERCISES OF CHAPTER 9
14. Solve ut = uxx , u(0, x) = ↵ sin x, u(t, 0) = u(t, ⇡) = 0.
Solution. We know that the solution satisfying u(t, 0) = u(t, ⇡) = 0
is given by
1
X
2
u(t, x) =
Ck e k t sin(kx).
k=1
The initial condition u(0, x) = ↵ sin x implies C1 = ↵ and Ck = 0 for
all k = 2, 3, .... Thus u(t, x) = ↵e t sin x.
15. Solve ut = c2 uxx , u(0, x) = ↵ sin x. with the boundary condition
u(t, 0) = u(t, ⇡) = 0.
Solution. Setting v(t, x) = u(c 2 t, x), we get vt (t, x) = c 2 ut (c 2 t, x) =
uxx (c 2 t, x) = vxx (t, x). Moreover,Pv(t, 0) = u(c 2 t, 0) = 0 and v(t, ⇡) =
2
u(c 2 t, ⇡) = 0. Then v(t, x) = 1
Ck e k t sin kx whereby u(t, x) =
1
P1
k 2 c2 t
sin kx. The initial condition u(0, x) = ↵ sin x implies
1 Ck e
2
C1 = ↵ and Ck = 0 for all k = 2, 3, ... Thus u(t, x) = ↵e c t sin x.
16. Solve ut = uxx , u(0, x) = ↵ sin
u(t, 0) = u(t, L) = 0.
⇡x
L
, with the boundary condition
Solution. Setting v(t, x) = u(t, Lx
), we get vt (t, x) = ut (t, Lx
) =
⇡
⇡
2
Lx
⇡
uxx (t, ⇡ ) = L2 vxx (t, x), Moreover, v(t, 0) = u(t, 0) = 0 and v(t, ⇡) =
u(t, L) = 0. Using the previous exercise with c = L⇡ , we find v(t, x) =
P1
k2 ⇡ 2 t
L2
C
e
sin kx, whereby
k
1
✓
◆
X
k2 ⇡ 2 t
k⇡
u(t, x) =
Ck e L2 sin
x .
L
Finally, the initial condition u(0, x) = ↵ sin
u(t, x) = ↵e
⇡2 t
L2
sin
⇡x
L
⇣ ⇡x ⌘
L
implies
.
Chapter 10
Solutions to exercises of
Chapter 10
1. Find x(t) =
P
ak tk such that tx00 = x.
Solution. One has
x00 (t) = 2a2 + 3 · 2 a3 t + 4 · 3 a4 t2 · · · =
X
k(k
1)ak tk
2
k 2
and hence
tx00 (t) = 2a2 t + 3 · 2 a3 t2 + 4 · 3 a4 t3 + · · · =
X
k(k
1)ak tk
k 2
The equation tx00 = x yields
2a2 t + 3 · 2 a3 t2 + 4 · 3 a4 t3 + · · · = a0 + a1 t + a2 t2 + · · ·
Hence a0 = 0 and 2a2 = a1 , 3 · 2 a3 = a2 , and so on. We deduce
a1
2
a2
a1
=
=
3·2
3 · 22
a3
a1
=
=
4·3
4 · 32 · 22
= ···
a1
=
2
k(k 1) (k 2)2 (k
a2 =
a3
a4
ak
135
3)2
· · · 22
=
a1
, k
k!(k 1)!
2.
1
136
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
Then the solution is the series, uniformly convergent on R,
a1 t 2
a1 t 3
a1 t k
+
+
·
·
·
+ ···
2
3 · 22
k!(k 1)!
X
tk
= a1
,
k!(k
1)!
k 1
x(t) = a1 t +
where it is understood that 0! = 1.
P
2. Find x(t) = k 0 ak tk such that tx00 = x0 .
P
P
k 1
00
Solution. Since x0 (t) =
ka
t
and
tx
(t)
=
k
k 1
k
1)ak tk 1 we find
X
X
k(k 1)ak tk 1 =
kak tk 1
k 2
2
k(k
(10.1)
k 1
namely
2a2 t + 3 · 2 a3 t2 + 4 · 3 a4 t3 + · · · = a1 + 2a2 t + 3a3 t2 + 4a4 t3 + · · ·
Then a1 = 0 and hence (10.1) becomes
X
X
k(k 1)ak tk 1 =
kak tk
k 2
whereby
X
(k(k
1)
k)ak tk
1
k 2
1
= 0 =) (k 2
k 2
2k)ak = 0, 8 k
2.
This implies that ak = 0 for all k 3, while a0 and a2 remain undetermined. Hence x(t) = a0 + a2 t2 .
P
3. Find x(t) = k 0 ak tk such that x00 = tx + 1 and x(0) = 0, x0 (0) = 1.
Solution. Since x(0) = a0 and x0 (0) = a1 it follows that a0 = 0, a1 = 1.
P
Now, let us solve the equation. Since tx = k 0 ak tk+1 one finds
X
k 2
k(k
1)ak tk
2
=
X
k 0
ak tk+1 + 1
137
that is
2a2 + 3 · 2 a3 t + 4 · 3 a4 t2 + · · · = 1 + t2 + a2 t3 + · · ·
In particular, a2 = 12 .
To find the general recursive formula, we have to notice that the coefficients of tk+1 in the two series are: (k + 3)(k + 2)ak+3 on the left hand
side, and ak on the right hand side. Thus we find
(k + 3)(k + 2)ak+3 = ak =) ak+3 =
or else
ak =
ak 3
,
k(k 1)
k
ak
(k + 3)(k + 2)
3.
Iterating the procedure, one gets
ak
ak 3
=
(k + 3)(k + 2)
(k + 3)(k + 2)k(k 1)
ak 6
=
= ···
(k + 3)(k + 2)k(k 1)(k 3)(k 4)
ak+3 =
It is convenient to distinguish among 3 cases:
1) k = 3n, n 1. One has (here and below it is understood that n is
such that all the factors in the denominators are positive)
a3n =
=
a3n
3n(3n
3n(3n
3
1)
a3n
1)(3n
6
3)(3n
4)
= ···
=
3n(3n
a0
1)(3n 3)(3n
4) · · · 3 · 2
=0
138
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
2) k = 3n + 1, n
1. One has
a3n 2
(3n + 1)3n
a3n
=
(3n + 1)3n(3n
a3n+1 = =
5
2)(3n
3)
= ···
a1
=
(3n + 1)3n(3n 2)(3n
1
=
(3n + 1)3n(3n 2)(3n
3) k = 3n + 2, n
3) · · · 4 · 3
3) · · · 4 · 3
.
1. One has
a3n 1
(3n + 2)(3n + 1)
a3n 4
=
(3n + 2)(3n + 1)(3n
a3n+2 =
1)(3n
= ···
a2
=
(3n + 2)(3n + 1)(3n 1)(3n
1
=
(3n + 2)(3n + 1)(3n 1)(3n
2)
2) · · · 5 · 4
2) · · · 5 · 4·2
.
In conclusion, the solution is given by
x(t) = t +
X
n 1
+
t3n+1
(3n + 1)3n(3n 2)(3n
3) · · · 4 · 3
t2 X
t3n+2
+
2 n 1 (3n + 2)(3n + 1)(3n 1)(3n
2) · · · 5 · 4 · 2
.
The first terms of the series are
✓
◆
t4
t7
t10
x(t) =
t+
+
+
+ ···
4 · 3 7 · 6 · 4 · 3 10 · 9 · 7 · 6 · 4 · 3
✓ 2
◆
t
t5
t8
t11
+
+
+
+
+ ··· .
2
5 · 4 · 2 8 · 7 · 5 · 4 · 2 11 · 10 · 8 · 7 · 5 · 4 · 2
139
4. Solve x00 + tx0 + x = 0, x(0) = 1, x0 (0) = 0.
P
Solution. Let x(t) = k 0 ak tk . The initial conditions imply a0 = 1
P
and a1 = 0. Thus x(t) = 1 + a2 t2 + a3 t3 + · · · = 1 + k 2 ak tk . One
P
P
k
00
has x0 (t) =
kak tk 1 and tx0 (t) =
k
2
k 2 kak t . Finally x (t) =
P
1)ak tk 2 . Then the equation yields
k 2 k(k
X
1)ak tk
k(k
2
+
k 2
If k
X
kak tk + 1 +
k 2
2 the coefficient of tk
2
ak t k = 0
k 2
is k(k
ak =
X
1)ak + (k
1)ak 2 , whereby
ak 2
.
k
If k is even, k = 2n, we find
a2n =
a2n 2
a2n
=
2n
2n(2n
4
2)
= · · · = ( 1)n
If k is odd, recalling that a1 = 0, we find a3 =
0, . . . , and in general ak = 0.
a0
n 1
=
(
1)
.
2n n!
2n n!
a1
3
= 0, a5 =
a3
5
=
Then the solution is given by
x(t) =
X
( 1)n
n 0
t2n
=1
2n n!
t2
t4
+
2
4 · 2!
t6
+ ···
8 · 3!
the series being uniformly convergent on all R.
5. Using the Frobenius method, solve 4t2 x00 + 4tx0
x = 0, t > 0.
x
Solution. Writing the equation in the equivalent form t2 x00 + tx0
=
4
1
0, we have p0 = 1 and q0 =
. The indicial equation is F (r) =
4
1
1
r(r 1) + r
= 0, namely r2
= 0 whose roots are r = ± 12 . Take
4
4
P
x(t) = t1/2 k 0 ak tk . Since
F
✓
1
+k
2
◆
=
✓
1
+k
2
◆2
1
= k 2 + k > 0,
4
8k
1
140
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
the recursive formula is
ak (k 2 + k) =
q 1 ak 1 ,
k
1.
Since all q1 = 0, it follows that ak = 0, for all k 1, while a0 remains
undetermined. Thus a first family of solutions is x1 (t) = ct1/2 , c a
constant.
If we take r =
1/2 we have that
✓
◆
1
F
+ k = k2
2
k > 0,
8k
2
This implies that ak = 0 for all k 2, while a0 , a1 are undetermined.
Then x(t) = t 1/2 (a0 + a1 t) = a0 t 1/2 + a1 t1/2 is a second family of
solutions, which contains, taking a0 = 0, the previous family x1 (t).
Since t 1/2 and t1/2 are linearly independent, the general solution is
x(t) = c1 t 1/2 + c2 t1/2 , c1 , c2 constants.
6. Using the Frobenius method, solve t2 x00 + 3tx0 = 0, t > 0.
Solution. Here the indicial equation is F (r) = r(r 1)+3r = 0, namely
r2 + 2r = 0, whose roots are r = 0 and r = 2. The root rP= 0 gives
the constant solution x(t) ⌘ c. Take r = 2 and x(t) = t 2 k 0 ak tk .
One has F ( 2 + k) = (k 2)2 + 2(k 2) = (k 2)k > 0 for all k 6= 0, 2.
Moreover p0 = q0 = q1 = 0 and hence the recursive formula yields
ak = 0 for all k 6= 0, 2. The coefficients a0 , a2 are undetermined. The
solution is x(t) = t 2 (a0 + a2 t2 ) = a0 t 2 + a2 .
7. Using the Frobenius method, solve t2 x00
3tx0 + (4
t)x = 0.
Solution. Since p0 = 3, q0 = 4, the indicial equation is F (r) =
r(r 1) 3r + 4 = r2 4r + 4 = (r 2)2 = 0, whose root is r = 2.
Since q1 = 1, the recurrence formula is F (2 + k)ak = ak 1 , namely
ak k 2 = ak 1 . Thus
a1 = a0 , a 2 =
a1
a0
a2
a0
a0
= 2 , a3 = 2 = 2 2 , ...; ak = 2
,
2
2
2
3
2 ·3
2 · · · k2
and hence
x(t) = a0 t2
X
tk
.
22 · · · k 2
141
Since r = 2 is a double root of the inditial equation,
a second solution,
P k+2
linearly independent of x is y(t) = x(t) ln t + bk t , where bk have
to be determined.
8. Using the Frobenius method, solve t2 x00 + tx0 + (t
1)x = 0.
Solution. The indicial equation is F (r) = r(r 1)+r 1 = r2 1 = 0,
whose solutions are r = ±1. If r = 1 then F (1 + k) = (1 + k)2 1 =
k(k + 2) and one has k(k + 2)ak = ak 1 . Then
a1 =
a0
, a2 =
3
a1
a0
=
, a3 =
2·4
2·3·4
a2
=
3·5
It is convenient to write these equations in the form
a0
a0
1
, a2 =
·
, a3 =
3
3 2·4
a1 =
and in general
ak =
(1)k
a0
, ...
2·3·4·3·5
a0
1
·
, ...
2
3 2·4·5
a0
1
·
.
k
1
3
2 · 4 · 5 · · · k · (k + 2)
The corresponding solution is
x(t) = a0 t
X
k 1
(1)k
tk
.
3k 1 · 2 · 4 · 5 · · · k · (k + 2)
Next, let us take r = 1. Now F ( 1 + k) = ( 1 + k)2
and hence for k = 1, 2 we find
a1 =
a0 ,
0 · a2 =
1 = k(k
a1 .
Then a0 = a1 = 0, a2 remains undetermined while for k
k(k 2)ak = ak 1 , namely
a3 =
a2
, a4 =
3
In general, for k
a3
a2
=
, a5 =
4·2
3·4·2
a4
=
5·3
a2
3 · 4 · 2 · · · k · (k
2)
,
(k
3 one has
a2
, ...
3·4·2·5·3
3, all ak depend on a2 and one has
ak = ( 1)k
2)
3).
142
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
The solution is
x(t) = t
1
a2 t 2 +
X
k
a2 t k
( 1)k
3 · 4 · 2 · · · k · (k
3
2)
!
.
9. Find the solution xa of the Bessel equation t2 x00 + tx0 + t2 x = 0 such
that xa (0) = a.
Solution. This a Bessel equation of order m = 0. The general solution
is x(t) = c1 J0 (t) + c2 Y0 (t). Since J0 (0) = 1 and Y0 (t) ! 1 as t ! 0,
the initial condition xa (0) = a implies c1 = a and c2 = 0. Thus the
solution is given by xa (t) = aJ0 (t).
10. Find the solution xa of the Bessel equation t2 x00 + tx0 + (t2
such that x0a (0) = a.
1)x = 0
Solution. The general solution is x(t) = c1 J1 (t) + c2 Y1 (t) and thus
x0 (t) = c1 J10 (t) + c2 Y10 (t). The condition x0a (0) = a implies c2 = 0
because Y10 (t) ! 1 as t ! 0. Since J10 (0) = 1 it follows that c1 = a
and the solution is given by xa (t) = aJ1 (t).
11. For all m > 0 show that t2 x00 + tx0 + (t2
solution such that x(0) = 0.
m2 )x = 0 has a nontrivial
Solution. The general solution is x(t) = c1 Jm (t) + c2 Ym (t). Since
Ym (t) ! 1 as t ! 0, we find c2 = 0 and hence x(t) = c1 Jm (t). For
t = 0 we find 0 = x(0) = c1 Jm (0). If m > 0 we have that Jm (0) = 0
and hence any x(t) = c1 Jm (t), c1 6= 0 is a non trivial solution.
12. Prove that (t2 J2 (t))0 = t2 J1 (t).
Solution. Since
J2 (t) =
X
k 0
( 1)k
k!(k + 2)!
✓ ◆2k+2
t
2
then
2
t J2 (t) =
X
k 0
( 1)k 2
t ·
k!(k + 2)!
✓ ◆2k+2 X
✓ ◆2k+4
t
t
( 1)k
2
=
2 ·
,
2
k!(k
+
2)!
2
k 0
143
the series being uniformly convergent on all R. Di↵erentiating each
term, one finds
2
(t J2 (t))
0
X ( 1)k 22 d ✓ t ◆2k+4
=
k!(k + 2)! dt 2
k 0
X ( 1)k 22 2k + 4 ✓ t ◆2k+3
=
k!(k + 2)!
2
2
k 0
✓ ◆2k+3
X ( 1)k 22
t
=
(k + 2)
k!(k + 2)!
2
k 0
Canceling (k + 2) we get
2
(t J2 (t))
0
X ( 1)k 22 ✓ t ◆2k+3
=
k!(k + 1)! 2
k 0
!
X ( 1)k ✓ t ◆2k+1
= t2
= t2 J1 (t).
k!(k
+
1)!
2
k 0
13. Show that J0 (t) has a strict maximum at t = 0.
✓ ◆2k
P
( 1)k t
. It follows that
Solution. We know that J0 (t) = k 0
(k!)2 2
J00 (0) = 0 and J000 (0) = 12 < 0 (recall that the series is uniformly
convergent and hence we can derive the series term by term).
J=
14. Let ↵ be a positive zero of J0 (t). Show that if J1 (↵) > 0 then J00 (↵) < 0.
Solution. Since (tJ1 (t))0 = tJ0 (t) then J1 (t) + tJ10 (t) = tJ0 (t). For
t = ↵ we find J1 (↵) + ↵J10 (↵) = 0. Then ↵J10 (↵) = J1 (↵). Since
↵ > 0 and J1 (↵) > 0, we infer that J10 (↵) < 0.
15. Setting Z(t) = J0 (t)
Z 0 (↵) = J00 (↵).
tJ1 (t), show that if ↵ is a zero of J0 (t) then
Solution. One has Z 0 (t) = J00 (t) (tJ1 (t))0 . Since (tJ1 (t))0 = tJ0 (t), we
infer that Z 0 (t) = J00 (t) tJ0 (t). Thus Z 0 (↵) = J00 (↵) ↵J0 (↵) = J00 (↵).
144
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
16. Using the power expansions of J0 , J1 , J2 , prove that J2 (t) =
J0 (t).
Solution. From J1 (t) =
P
k 0
( 1)k
k!(k + 1)!
2
J1 (t)
t
✓ ◆2k+1
t
it follows
2
X ( 1)k ✓ t ◆2k
2
J1 (t) =
t
k!(k + 1)! 2
k 0
✓ ◆2
✓ ◆4
1 t
1
t
= 1
+
2! 2
2! · 3! 2
1
3! · 4!
✓ ◆6
t
+ ···
2
Since
✓ ◆2
✓ ◆4
X ( 1)k ✓ t ◆2k
t
1
t
J0 (t) =
=1
+
2
2
(k!)
2
2
(2!) 2
k 0
1
(3!)2
✓ ◆6
t
+· · ·
2
then
2
J1 (t)
t
J0 (t) =
X
( 1)
k
( 1)
k
k 0
=
X
✓
✓
1
k!(k + 1)!
1
(k!)2 (k
+ 1)
✓ ◆2k
X
k
t
k
=
( 1)
k!(k + 1)! 2
k 0
k 0
◆ ✓ ◆2k
t
2
◆ ✓ ◆2k
1
t
2
(k!)
2
1
(k!)2
Relabeling the index by setting k = h + 1, we find
X
k
k
( 1)
k!(k + 1)!
0
For h =
k
✓ ◆2k
t
=
2
✓ ◆2h+2
t
( 1)
2
h
1
✓ ◆2h+2
X
h+1
t
h+2
=
( 1)
(h + 1)!(h + 2)! 2
h
1
1 the coefficient of
X
t 2h+2
2
h+1
(h + 1)
(h + 1)!(h + 2)!
is zero. Hence, simplifying h + 1,
145
we have
X
k
k
( 1)
k!(k + 1)!
0
k
✓ ◆2k
✓ ◆2h+2
X
t
1
t
h+2
=
( 1)
2
h!(h + 2)! 2
h 0
✓ ◆2h+2
X
1
t
h
=
( 1)
h!(h + 2)! 2
h 0
which is nothing but J2 (t).
0
17. Prove that Jm
(t) = Jm 1 (t)
P
Solution. Since Jm = k 0
0
Jm
(t)
=
=
=
=
=
=
Jm
m
Jm (t), m an integer.
t
( 1)k
k!(k+m)!
t 2k+m
2
one has
X ( 1)k (2k + m) ✓ t ◆2k+m 1
1 (t) =
2k!(k + m)!
2
k 0
✓ ◆2k+m
X
( 1)k
t
k!(k + m 1)! 2
k 0
✓ ◆2k+m
1
t
( 1)
k!(k + m 1)!
2
k 0

✓ ◆2k+m 1
X
2(k + m)
t
k 2k + m
( 1)
2k!(k + m)!
2
k 0

✓
◆
2k+m 1
X
m
t
k
( 1)
2k!(k + m)!
2
k 0

✓
◆ 1 ✓ ◆2k+m
X
m
t
t
k
( 1)
2k!(k + m)!
2
2
k 0

✓ ◆2k+m
X
m
2 t
k
( 1)
2k!(k + m)! t 2
k 0
✓ ◆2k+m
m X ( 1)k
t
m
=
Jm (t).
t k 0 k!(k + m)! 2
t
X
k

2k + m
2k!(k + m)!
1
1
146
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
18. Prove if ↵1 is the first positive zero of J0 (t), then J1 (↵1 ) > 0 and
J2 (↵1 ) > 0.
Solution. We know that between two consecutive zeros of J1 there is
a zero of J0 . Since ↵1 is the first positive zero of J0 , it follows that the
the first positive zero of J1 is greater than ↵1 and hence J1 (↵1 ) > 0.
2
Moreover, from J2 (t) = J1 (t) J0 (t) (see Exercise n.15) it follows
t
2
that J2 (↵1 ) =
J1 (↵1 ) > 0.
↵1
19. Let ↵1 denote the first positive zero of J0 . Show that the only solution
of t2 x00 + tx0 + (t2 1)x = 0 such that x(0) = x(↵1 ) = 0 is x(t) ⌘ 0.
Solution. The solution satisfying the condition x(0) = 0 is x(t) =
c1 J1 (t). Since J1 (↵1 ) > 0 (see the preceding exercise) the condition
x1 (↵1 ) = 0 implies c1 = 0.
20. Find
> 0 such that the problem
s
d2 y
+ y(s) = 0,
ds2
y(0) = 0,
y(1) = 0,
(*)
has a nontrivial solution.
p
p
p
Solution. Setting t = 2 s and y(s) = 2 s · x(2 s), we know that
2
s ddsy2 + y = 0 is transformed into the Bessel equation of order 1
t2 x00 (t) + tx0 (t) + (t2
1)x(t) = 0
Take the family of solutions x(t) = cJ1 (t) which yields
p
p
y(s) = 2c s J1 (2 s)
Notice that for s =p0 we have
the condition
p y(0) = 0, as required. Next,
⇤
y(1) = 0 yields 2c
J1 (2 ) = 0. It follows that ( ) has nontrivial
✓ ◆2
p
p
n
,
solutions given by yn (s) = 2c n sJ1 (2 n s), provided n =
2
n being the positive zeros of J1 .
21. Find the positive integer
such that t2 x00 + tx0 + t2 x =
nontrivial solution satisfying x(0) = 0, x0 (0) = 1.
x has a
147
Solution. Setting = m2 the given equation becomes t2 x00 +tx0 +(t2
m2 )x = 0, which is the Bessel equation of order m. The general solution
is x = c1 Jm +c2 Ym . Recall that the properties of Bessel function Jm , Ym
when m is not an integer are the same that those when m 2 Z. The
condition x(0) = 0 implies c2 = 0 and Jm (0) = 0 and thus m 6= 0. Since
0
0
x0 (t) = c1 Jm
(t) it follows that 1 = x0 (0) = c1 Jm
(0). If m 6= 1 we know
0
that Jm (0) = 0. Thus m = 1. In conclusion, the only nonnegative
integer satisfying the statement is = 1.
148
CHAPTER 10. SOLUTIONS TO EXERCISES OF CHAPTER 10
Chapter 11
Solutions to exercises of
Chapter 11
1. Find the L-transform of sinh !t = 12 (e!t
e !t ).
Solution. Since L is linear, one has
L{sinh !t} =
1
2
!t
} =
1
2
✓
} =
1
2
✓
L{e
!t
L{e } + L{e
!t
L{e }
e
!t
) and cosh !t = 12 (e!t +
1
s+!
◆
=
1
+
! s+!
◆
=
1
s
!
!
s2
!2
and
L{cosh !t} =
1
2
!t
1
s
s
s2
!2
2. Find the L-transform of t sin !t and t cos !t.
Solution. We use property (P 5) with n = 1. Then
L{t sin !t}(s) =
d
L{sin !t}(s) =
ds
d
ds
✓
d
L{cos !t}(s) =
ds
d
ds
✓
!
2
s + !2
◆
=
◆
=
(s2
2!s
.
+ ! 2 )2
Similarly
L{t cos !t}(s) =
149
s
2
s + !2
s2 ! 2
.
(s2 + ! 2 )2
.
150
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
3. Find the L-transform of t sinh !t and t cosh !t.
Solution. Property (P 5) with n = 1 yields
✓
◆
d
d
!
2!s
L{t sinh !t} =
L{sinh !t}(s) =
= 2
.
2
2
ds
ds s
!
(s
! 2 )2
Similarly
d
L{cosh !t}(s) =
ds
L{t cosh !t} =
d
ds
✓
s
s2
!2
◆
=
s2 + ! 2
.
(s2 ! 2 )2
4. Find the L-transform of e↵t sin !t and e↵t cos !t.
Solution. Using property (P 1) we find
L{e↵t sin !t}(s) = L{sin !t}(s
↵) =
L{e↵t cos !t}(s) = L{cos !t}(s
↵) =
and
(s
!
.
↵)2 + ! 2
s ↵
.
(s ↵)2 + ! 2
5. Find L{f } where f (t) = 1, if 0  t  1, f (t) = 2, if 3  t  4 and
f (t) = 0 otherwise.
Solution. The function f can be written as f (t) =
Then
L{f } = L{ [0,1] } + 2L{ [3,4] } =
=
1
e
s
s
+2
e
3s
e
s
4s
=
1
e
s
+ 2e
s
[0,1] (t) + 2 [3,4] (t).
3s
6. Find the L-transform of t ⇤ et .
Solution. One has
L{t ⇤ et } = L{t} · L{et } =
1
1
·
.
2
s s 1
7. Find the L-transform of t2 ⇤ eat .
Solution. One has
L{t2 ⇤ eat } = L{t2 } · L{eat } =
2
1
·
.
3
s s a
2e
4s
.
151
8. Let f be a piecewise continuous T -periodic function. Show that L{f }
exists and
Z T
1
L{f } =
·
e st f (t)dt.
1 e sT
0
Solution. Using the definition we have
Z +1
Z T
Z
st
st
L{f } =
e f (t)dt =
e f (t)dt +
0
0
+1 Z
X
=
0
2T
e
st
T
f (t)dt + · · ·
(n+1)T
st
e
f (t)dt,
nT
provided the series converges. The change of variable t = ⌧ + nT yields
Z
Z
(n+1)T
e
st
f (t)dt =
nT
T
s(⌧ +nT )
e
f (⌧ + nT )d⌧.
0
Using the periodicity of f it follows
Z
(n+1)T
st
e
f (t)dt =
nT
Z
T
e
L{f } =
Notice that
1
sum is
1 e
0
sT
e
nT s
f (⌧ )d⌧ = e
nT s
0
Thus
P+1
s(⌧ +nT )
+1
X
e
nT s
0
·
Z
Z
T
e
s⌧
T
e
s⌧
f (⌧ )d⌧.
0
is the geometric series with ratio e
. Then
L{f } =
1
1
e
sT
·
Z
T
e
s⌧
f (⌧ )d⌧
0
as claimed.
9. Find the L-transform of the 2-periodic square wave function
f (t) = 1, if 0  t < 1,
and f (t + 2) = f (t) for all t
2.
f (⌧ )d⌧.
0
f (t) = 0, if 1  t < 2
sT
whose
152
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
1
2
1
3
4
Figure 11.1: the square wave function
Solution. Here
1
L{f } =
1 e
2s
·
1
=
1 e
Z
2
e
s⌧
0
2s
1
f (⌧ )d⌧ =
1 e

s
e
· 1
=
s
s
s(1
2s
e
e
·
Z
1
e
s⌧
d⌧ =
0
s
2s )
.
10. Find the L-transform of the saw-tooth T-periodic function
f (t) = t, if 0  t < T ,
f (t + T ) = f (t).
Solution. One has
L{f } =
1
1
e
sT
·
Z
T
e
s⌧
⌧ d⌧.
0
Integrating by parts we find
Z
Z
e s⌧ ⌧
e s⌧
e s⌧ ⌧
e s⌧
s⌧
e ⌧ d⌧ =
+
d⌧ =
=
s
s
s
s2
✓
◆
e s⌧
1
(s⌧ + 1)e s⌧
=
· ⌧+
=
.
s
s
s2
153
T
3T
2T
T
Figure 11.2: the saw-tooth function
and therefore
Z T
e s⌧ ⌧ d⌧ =
(sT + 1)e
s2
0
sT
+
1
1
=
2
s
(sT + 1)e
s2
sT
.
In conclusion,
L{f } =
1
(sT + 1)e sT )
1
= 2
2
sT
s (1 e )
s
T e sT
.
s(1 e sT )
11. Let F (s) = L{f }(s) be defined for s > 0. Suppose that |f (t)|  C for
all t 0. Show that lims!+1 F (s) = 0.
Solution. One has
|F (s)| 
Since
R +1
0
e
st
Z
+1
e
0
st
|f (t)|dt  C
Z
+1
e
st
dt.
0
1
C
dt = L{1} = , then |F (s)| 
and the result follows.
s
s
12. Let F (s) = L{f }(s) be defined for s > 0. Suppose that f (t)
for all t 0. Show that lims!0+ F (s) = +1.
Solution. One has
F (s) =
Z
+1
e
0
st
f (t)dt
C
Z
+1
e
0
st
dt =
C
.
s
C>0
154
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
Thus,
C
= +1.
s!0+ s
lim F (s)
lim
s!0+
13. Find the inverse L-transform of
2
and
s
.
4
4
Solution. Using exercise 1, with ! = 2, we find
⇢
⇢
2
s
1
1
L
= sinh 2t, L
= cosh 2t.
2
2
s
4
s
4
s2
s2
1
s 1
and 2
.
2s + 2
s
2s + 2
2s + 2 = 1 + (s 1)2 , we can use exercise 4 with
14. Find the inverse L-transform of
Solution. Since s2
↵ = ! = 1 yielding
⇢
1
1
L
=e
2
s
2s + 2
⇢
1
1
15. Find L
.
s2 3s + 2
Solution. Since s2
s2
Then
L
1
⇢
s2
t
L
sin t,
3s + 2 = (s
1
=
3s + 2
(s
1
3s + 2
s2
=
L
⇢
2)
1
s
⇢
1)(s
1
1)(s
1
1
1
s
s2
1
2s + 2
=e
t
cos t.
2) we have
=
+L
1
s
1
1
⇢
+
1
s
1
s
2
2
=
.
et + e2t .
The same result could be found by using Theorem 11.3.7 with P (s) = 1
and Q(s) = s2 3s + 2.
⇢
s 2
1
16. Find L
.
s3 s
Solution. Let us use Theorem 11.3.7 with P (s) = s 2 and Q(s) =
s3 s. The roots of Q = 0 are 1 = 0, 2 = 1, 3 = 1. Since
Q0 (s) = 3s2 1 then Q0 (0) = 1, Q0 (±1) = 2 and hence
⇢
s 2
P (1) t P ( 1) t
1 t 3 t
1
L
= P (0)e0 +
e +
e =2
e
e .
3
s
s
2
2
2
2
155
17. Find L
1
⇢
1
s4
1
Solution. One has
1
s4
and thus
⇢
1
1
L
4
s
1
1
=
.
1
1
=
2
2
1)(s + 1)
2(s
1)
(s2
1
= L
2
1
⇢
1
s2
1
1
L
2
1
⇢
1
+ 1)
2(s2
1
+1
s2
=
1
sinh t
2
1
sin t.
2
⇢
s2 + 3s + 1
.
s2 + s
Solution. From
18. Find L
1
s2 + 3s + 1
1
1
=1+ +
2
s +s
s s+1
it follows
⇢ 2
s + 3s + 1
1
L
s2 + s
=L
1
{1}+L
1
⇢
1
+L
s
1
⇢
1
s+1
= +1+e t .
19. Let F (s) = L{f }. Show that if f (t) > 0 then F is decreasing and
concave upward.
Solution. Using property (P 5) we find
Z
0
F (s) = L{tf (t)} =
00
2
F (s) = L{t f (t)} =
Z
+1
e
st
tf (t)dt,
0
+1
e
st 2
t f (t)dt.
0
Since f (t) > 0 it follows that both the integrals in right hand side are
positive. Thus F 0 (s) < 0 and F 00 (s) > 0.
Rt
20. Prove property (P 3) of the L-transform: if g(t) = 0 f (⌧ )d⌧ then
L{g}(s) = L{fs}(t) .
Solution. Since g 0 (t) = f (t), the result follows by applying property
(P 4).
156
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
21. Use (P 3) to show that L{t} = s 2 .
Solution. As in the preceding exercise with g(t) = t, f (t) = 1. Then
L{t} =
L{1}
1
= 2.
s
s
22. Use (P 4) to find the L-transform of e↵t .
Solution. Property (P 4) with n = 1 yields L{f 0 (t)} = sL{f (t)}
f (0). Then, taking f = e↵t , it follows L{↵e↵t )} = sL{e↵t } 1. Since
L{↵e↵t } = ↵L{e↵t } we have ↵L{e↵t } = sL{e↵t } 1 and thus
L{e↵t }(s) =
1
s
↵
,
(s > ↵).
23. Let J0 (t) be the Bessel function of order 0 satisfying tx00 + x0 + tx = 0,
J0 (0) = 1, J00 (0) = 0. Find X (s) = L{J0 (t)} such that X (0) = 1.
Solution. From tJ000 + J00 + tJ0 = 0 we infer
L{tJ000 } + L{J00 } + L{tJ0 } = 0.
(a)
Let us evaluate the three terms separately. Using (P 5) with n = 1, it
follows
d
L{tJ0 } =
L{J0 } = X 0 (s)
(b)
ds
and
d
L{tJ000 } =
L{J000 }.
ds
Using (P 4) we find
L{J000 } = s2 L{J0 (t)}
sJ0 (0)
J00 (0) = s2 X
s,
whereby
L{tJ000 } =
d 2
(s X
ds
s) =
2sX (s)
s2 X 0 (s).
(c)
Next, using again (P4),
L{J00 }(s) = sL{J0 (t)}
J0 (0) = sX (s)
1.
(d)
157
Substituting (b), (c), (d) into (a) it follows that X verifies
(1 + s2 )X 0 (s) + sX (s) = 0,
which is a separable equation. Integrating one finds
ln |X (s)| =
1
2
ln(1 + s2 ) + c.
From X (0) = 1 it follows that c = 0 and hence
ln |X (s)| =
1
2
ln(1 + s2 ) = ln p
1
.
1 + s2
In particular, X (s) does not vanish. Using again the fact that X (0) = 1,
we infer that X (s) > 0 and thus
X (s) = p
1
.
1 + s2
24. Solve x0 + x = et , x(0) = 0 using the L-transform.
Solution. By linearity, L{x0 } + L{x} = L{et }. Letting X(s) = L{x}
and using (P 4) one has
sX(s)
x(0) + X(s) = L{et } =
1
s
1
.
Since x(0) = 0 we infer
(s + 1)X(s) =
1
s
1
whereby
1
X(s) =
(s + 1)(s
1
=
1)
2
✓
Taking the inverse L-transform one finds

⇢
⇢
1
1
1
1
1
x(t) =
L
L
2
s 1
s+1
1
s
1
1
s+1
1
= (et
2
◆
e t ) = sinh t.
158
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
25. Solve x0 + x = t, x(0) =
1 using the L-transform.
Solution. Arguing as before, we let X = L{x} finding
(s + 1)X(s) + 1 = L{t} =
1
1
=) (s + 1)X(s) = 2
2
s
s
and thus
1
1
.
s2
s2 s
Taking the inverse L-trasform, we obtain x(t) = t
X(s) =
26. Solve x00
1
s
1=
s2
1
s2
=
1.
2x0 + x = 0, x(0) = x0 (0) = 1 using the L-transform.
Solution. Letting X = L{x} and taking the L-transform one finds
s2 X
x0 (0)
sx(0)
2(sX
x(0)) + X = 0.
Using the initial values it follows
s2 X
s
1
2(sX
1) + X = 0.
Solving with respect to X we infer
X=
s
s2
1
s
=
2s + 1
(s
Then
1
x(t) = L {X} = L
27. Solve x00
1
1
1
=
.
2
1)
s 1
⇢
1
s
1
= et .
4x0 + 3x = 1, x(0) = x0 (0) = 0 using the L-transform.
Solution. Letting X = L{x} and taking the L-transform one finds
s2 X
sx(0)
x0 (0)
4(sX
1
x(0)) + 3X = L{1} = .
s
Using the initial values it follows
s2 X
whereby
X=
s(s2
1
4sX + 3X = .
s
1
=
4s + 3)
s(s
1
3)(s
1)
.
159
Splitting
1
3)(s
s(s
1)
into partial fractions,
1
3)(s
s(s
1)
=
a
b
c
+
+
s s 3 s 1
we find
1 = a(s2 4s+3)+bs(s 1)+cs(s 3) = (a+b+c)s2 (4a+b+3c)s+3a.
Then
a + b + c = 0,
4a + b + 3c = 0,
1
3
a=
whereby
b+c=
1
,
3
b + 3c =
Thus
X=
4
3
1
b= ,
6
=)
1
1
+
3s 6(s 3)
c=
1
.
2
1
2(s
1)
and hence
1
x(t) = L {X} = L
3t
=
1 e
+
3
6
1
t
e
.
2
⇢
1
3s
+L
1
⇢
1
6(s
3)
L
1
⇢
1
2(s
1)
The same result could be found by using Theorem 11.3.7 with P (s) = 1
and Q(s) = s(s 3)(s 1).
28. Solve x0000 + x00 = 0, x(0) = 0, x0 (0) = 1, x00 (0) = x000 (0) = 0 using the
L-transform.
Solution. Letting X = L{x} and taking the L-transform one finds
s4 X
s3 x(0)
s2 x0 (0)
sx00 (0)
x000 (0) + s2 X
sx(0)
x0 (0) = 0
Using the initial values it follows
s4 X
s2 + s2 X
1 = 0 =) (s4 + s2 )X = s2 + 1
160
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
Dividing by s4 + s2 = s2 (s2 + 1) one finds
X=
s2 + 1
s2 + 1
1
=
= 2.
4
2
2
2
s +s
s (s + 1)
s
Thus
1
⇢
1
x = L {X} = L
29. Find the ”generalized” solution of x0
1
s2
= t.
x = Ha (t), x(0) = 0.
Solution. Letting X = L{x} and taking the L-transform one has
sX
X=
If we write F (s) =
as
e
=)
s
1
,
s 1
e
s(s
X=
G(s) =
as
e
s
as
1)
=
1
s
1
·
as
e
s
we find X = F (s)G(s) and hence
x(t) = L 1 {X} = L 1 {F G} = L 1 {F } ⇤ L 1 {G}.
Since
1
L {F } = L
1
⇢
1
s
t
=e,
1
L {G} = L
we deduce
t
1
x(t) = e ⇤ Ha (t) =
Z
1
⇢
as
e
s
= Ha (t),
t
et ✓ Ha (✓)d✓.
0
Since Ha (✓) = 0 for ✓ < a and = 1 for ✓ a, we infer
8
if t < a
< 0
x(t) =
: Rt t ✓
e d✓ = 1 + et a if t a,
a
which solves the equation for all t 6= a, while x0 (a ) = 0, x0 (a+) = 1.
30. Find the ”generalized” solution of x0 + x = Ha (t), x(0) = 0.
Solution. Repeating the previous calculations, we let X = L{x} finding
1
e as
X=
·
s+1
s
161
and x(t) = (f ⇤ g)(t) with f (t) = e t and g(t) = Ha (t). Then
8
Z t
if t < a
< 0
(t ✓)
x(t) =
e
Ha (✓)d✓ =
: R t t+✓
0
e
d✓ = 1 ea t if t a.
a
As in the previous exercise, x solves the equation for all t 6= a, while
x0 (a ) = 0, x0 (a+) = 1.
t-a
-1+e
1-ea-t
a
a
(a) Solution of x0
x = Ha (t), x(0) = 0. (b) Solution of x0 + x = Ha (t), x(0) = 0.
31. Find the ”generalized” solution of x0
x = k , x(0) = a, t
0.
Solution. Letting X = L{x} and taking the L-transform one finds
sX
a
X = L{k } = k
=)
Then x(t) = (k + a)et , which solves x00
condition x(0) = k + a.
X(s) =
k+a
s 1
x = 0 with the new initial
32. Find the ”generalized solutions” of x00 + x = g(t) , x(0) = x0 (0) = 0,
where g is any piecewise continuous function with L-transform G(s)
defined for s > 0. In particular, solve in the case that g(t) = [0,1] (t)
Solution. Letting X = L{x} we find
s2 X + X = L{g} = G(s)
and hence
X=
G(s)
.
1 + s2
162
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
1
we infer L 1 {F } = sin t. Then taking the inverse
1 + s2
L-transform we deduce
Setting F (s) =
x(t) = L 1 {X} = L 1 {F G} = L 1 {F } ⇤ L 1 {G} = sin t ⇤ g(t).
If g(t) =
[0,1] (t)
we find
x(t) =
Z
t
sin(t
✓)
[0,1] (✓)d✓.
0
= 1 for 0  ✓  1 and [0,1] (✓) = 0 for ✓ > 1, it follows
Z t
Z t
x(t) =
sin(t ✓) [0,1] (✓)d✓ =
sin(t ✓)d✓ = 1 cos t, if t 2 [0, 1],
Since
[0,1] (✓)
0
0
while
x(t) =
=
Z
Z
1
sin(t
✓)
[0,1] (✓)d✓
+
0
Z
t
sin(t
✓)
[0,1] (✓)d✓
=
1
1
sin(t
✓)d✓ = cos(t
1)
cos t,
if t > 1.
0
Notice that x(t) = 1 cos t satisfies x00 + x = 1 in [0, 1[ with the initial
conditions x(0) = x0 (0) = 0, while x(t) = cos(t 1) cos t satisfies
x00 + x = 0 in ]1, +1) with the initial conditions x(1) = 1 cos 1,
x0 (1) = sin 1. Moreover, x(t) is di↵erentiable for all t
0, twice
00
di↵erentiable for t 6= 1 and x (t) has a simple discontinuity at t = 1
because x00 (1 ) = cos 1, x00 (1+) = cos 1 1.
33. Find the ”generalized” solution of x00 = (t
where a > 0.
a), x(0) = 1, x0 (0) = 0,
Solution. Letting X = L{x} we find
s2 X
Then
X=
s = eas .
s + eas
1 eas
=
+ 2
s2
s
s
whereby
x(t) = 1 + L
1
⇢
eas
s2
.
163
x
cos(t-1)-cos t
1
t
1-cos t
Figure 11.3: Solution of x00 + x =
Property (P 100 ) yields
⇢ as
e
1
L
= Ha (t)L
s2
[0,1] ,
⇢
1
s2
(t
a) =
⇢
1
1+t
1
x(0) = x0 (0) = 0
a) = Ha (t)(t
a).
In conclusion, we find
x(t) = 1 + Ha (t)(t
if 0  t < a
a if t a
Let us remark that x0 is discontinuous at t = a, because x0 (t) = 0 for
t < a, while x0 (t) = 1 for t > a.
34. Solve x(t) = 1 + e2t ⇤ x(t)
Solution. Letting X = L{x} we have
X = L{1} + L{e2t } · X.
Thus
X=
namely
✓
1
1
1
+
·X
s s 2
1
s
2
◆
1
X= .
s
164
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
x
1
a
t
Figure 11.4: Solution of x00 = (t
Then
X=
a), x(0) = 1, x0 (0) = 0, a > 0
s 2
2
1
=
+
s(s 3)
3s 3(s 3)
and hence
⇢
2
1
{X} = L
+
3s 3(s 3)
⇢
⇢
2 1 1
1 1
1
2 1
= L
+ L
= + e3t .
3
s
3
s 3
3 3
x=L
1
1
The same result could be found by using Theorem 11.3.7 with P (s) =
s 2 and Q(s) = s(s 3).
35. Solve x(t) = t3 + sin t ⇤ x(t).
Solution. Letting X = L{x} we have
X = L{t3 } + L{sin t} · X =
3!
1
+
· X.
s4 1 + s2
It follows
✓
1
1
1 + s2
◆
X=
6
s2
6
=)
·X = 4
4
2
s
1+s
s
165
and hence
6(1 + s2 )
6
6
= 6 + 4.
6
s
s
s
Taking the inverse L-transform we find
✓ 5
◆
t
t3
t5
x(t) = 6
+
=
+ t3 .
5! 3!
20
X=
36. Solve x0
k ⇤ x = 1, x(0) = 0, k > 0 a constant.
Solution. Let X = L{x}. From L{x0 }
sX
x(0)
L{k ⇤ x} = L{1} it follows
1
L{k} · X = ,
s
namely
kX
1
= .
s
s
Taking into account that x(0) = 0 we infer
sX
kX
1
= ,
s
s
sX
Then
x(t) = L
x(0)
1
⇢
i.e. X =
1
1
(
1
s2
k
.
p )
1
k
p · 2
=
k
k s
=L
s2 k
( p )
⇣p ⌘
1
k
1
p
= p ·L 1
=
·
sinh
kt .
s2 k
k
k
37. Solve x0 + (k 2 ) ⇤ x = 1, x(0) = 1, k > 0.
Solution. Letting X = L{x} we find
sX
1+
k2X
1
= ,
s
s
i.e. X =
s+1
.
s2 + k 2
To evaluate the inverse L-transform of X we notice that
s+1
s
1
= 2
+ 2
.
2
2
2
s +k
s +k
s + k2
166
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
Then
1
x(t) = L
=L
1
⇢
{X} = L
s
2
s + k2
1
+L
1
⇢
⇢
s2
s
1
+ 2
2
+k
s + k2
1
+ k2
s2
= cos k t +
=
sin k t
.
k
38. Solve the system
⇢
x0 = 2x + y,
y 0 = x 4y,
x(0) = 0
y(0) = 1
Solution. Taking the L-transform we find
⇢
sX
= 2X + Y
sY 1 = X 4Y
where X = L{x} and Y = L{y}.
Then
X=
Y
s
2
and
sY
1=
Y
s
4Y,
2
=)
s(s
2)Y + Y + 4(s
2)Y = s
2
namely
(s2 + 2s
7)Y = s
2,
=)
Y =
s2
s 2
.
+ 2s 7
To find the inverse L-transform of Y we can use Theorem 11.3.7,
p with
P = s 2 and Q = s2 + 2s 7, whose roots are 1,2 = 1 ± 8. Thus
y = A1 e
1t
+ A2 e
2t
Using the second equation x =
x=
= (4
1 A1 e
1t
1 )A1 e
,
Ai =
i
2
.
i+2
y 0 + 4y we find
2 A2 e
1t
P ( i)
=
Q0 ( i )
2
+ (4
2t
+ 4 A1 e
2 )A2 e
2t
1t
.
+ A2 e
2t
167
39. Solve the system
⇢
x0 = x + y,
y 0 = x + y,
x(0) = 1
y(0) = 0
Solution. Taking the L-transform and setting X = L{x}, Y = L{y},
we find
⇢
⇢
sX 1 = X + Y,
(s + 1)X = 1 + Y
sY
= X + Y,
(s 1)Y = X
Solving this algebraic system we get
X=
1+Y
,
s+1
(s
1)Y =
1+Y
s+1
Thus
s2
1 Y =1+Y
and hence
(s2
2)Y = 1 i.e. Y =
1
s2
2
.
Therefore
y=L
1
⇢
1
s2
2
1
= p ·L
2
1
( p
s2
2
2
)
p
sinh 2 t
p
=
.
2
From the second equation of the di↵erential system we find
p
p
sinh 2 t
0
p
x = y y = cosh 2 t
.
2
40. Solve the system
⇢
x0 = x + y,
y0 = y + ,
x(0) = 0
y(0) = 0
Solution. Taking the L-transform and setting X = L{x}, Y = L{y},
we find
⇢
⇢
sX = X + Y
(s 1)X = Y
=)
sY = Y + 1
(s + 1)Y = 1
168
CHAPTER 11. SOLUTIONS TO EXERCISES OF CHAPTER 11
Solving, it follows
Y =
1
,
s+1
X=
Y
s
1
=
1
s2
and hence
⇢
1
= e t,
s+1
⇢
1
x=L 1
= sinh t.
2
s
1
y=L
1
Notice that x, y solve the system
⇢ 0
x = x + y,
y 0 = y,
with the new initial conditions x(0) = 0, y(0) = 1.
1
Chapter 12
Solutions to exercises of
Chapter 12
1. Show that (0, 0) is asymptotically stable for the linear system
⇢
x0 =
2x + y
0
y = 7x 4y
Solution. The coefficient matrix is A =
are the solutions of
2
1
✓
2
7
1
4
◆
. Its eigenvalues
= 0, namely ( +2)( +4) 7 =
7
4
p
2
+6 +1 = 0. Then = 3± 8. Both these eigenvalues are negative
and hence (0, 0) is asymptotically stable.
2. Show that (0, 0) is asymptotically stable for the linear system
⇢
x0 =
x
y 0 = 4x
y
y
✓
1
4
Solution. The eigenvalues of A =
1
1
1
◆
are the solutions of
1
= 0, namely ( + 1)2 + 4 = 0. Then = 1 ± 2i.
4
1
They have negative real part and hence (0, 0) is asymptotically stable.
169
170
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
3. Show that (0, 0) is stable for the linear system
⇢ 0
x = x y
y 0 = 3x y
Solution. The eigenvalues of A =
1
1
✓
1
3
1
1
◆
are the solutions of
= 0, namely ( 2 1) + 3 = 2 + 2 = 0. Then
3
1
p
= ±i 2. They are purely imaginary and hence (0, 0) is stable.
4. Study the stability of (0, 0) for the system
⇢ 0
x =
2ax y
0
y = (9 + a2 )x
depending on the parameter a.
Solution. The eigenvalues of A =
✓
2a
9 + a2
1
0
◆
2a
1
= 0, namely ( +2a)+9+a2 =
2
9+a
whereby = a ± 3i. Then:
are the solutions of
2
+2a +9+a2 = 0
(i) if a < 0 then (0, 0) is unstable;
(ii) if a > 0 then (0, 0) is asymptotically stable;
(iii) if a = 0 then (0, 0) is stable, but not asymptotically stable.
5. Show that (0, 0) is unstable for the system
⇢ 0
x =
x + 4y
0
y = 3x 5y
and find the stable and unstable manifold.
✓
◆
1
4
Solution. The eigenvalues of the matrix A =
are the
3
5
solutions of ( + 1)( + 5) 12 = 2 + 6✓ 7◆= 0, that is 1 = 7, 2 =
a1
1. Thus (0, 0) is unstable. If e1 =
is any nontrivial vector
b1
such that Ae1 = 1 e1 = 7e1 , the stable manifold is the linear space
171
spanned by e1 , namely the straight line x = a1 t, y = b1 t, or equivalently
b1
y = x. The system Ae1 = 7e1 becomes
a1
⇢
a1 + 4b1 = 7a1
3a1 5b1 = 7b1
✓
◆
2
Solving, we find (up to constants) e1 =
and hence the stable
3
3
manifold is y =
x.
2
Similarly, solving
⇢
a2 + 4b2 = a2
3a2 5b2 = b2
✓ ◆
2
we find (up to constants) e2 =
. Thus the unstable manifold is
1
b2
1
y = x = x.
a2
2
6. Study the stability of the trivial solution of the equation x00 +2x0 x = 0.
2
Solution. The characteristic
+2
1 = 0, whose
p equation is
solutions are 1,2 = 1 ± 2. One of them is positive and one is
negative. Then x = 0 is unstable.
7. Study the stability of the trivial solution of the equation x00 +2x0 +x = 0.
Solution. The characteristic equation is 2 + 2 + 1 = 0 which has
the double root = 1 < 0. Thus x = 0 is asymptotically stable.
8. Study the stability of the trivial solution of the equation x00 + 2hx0 +
k 2 x = 0, h, k 6= 0.
Solution. The characteristic
equation is
p
2
solutions are = h ± h
k2.
2
+ 2h + k 2 = 0 whose
Case 1): h2 k 2 0. The eigenvalues are real. Moreover, if h < 0 both
the eigenvalues are positive and hence the trivial solution is unstable.
If h > 0 both the eigenvalues are negative and hence the trivial solution
is asymptotically stable.
Case 2): h2 k 2 < 0. The eigenvalues are complex conjugated with
real part h. Thus if h < 0 we have instability while if h > 0 we have
asymptotic stability.
172
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
9. Show that the equilibrium of the system
8 0
2x1 + x2 + x3
< x1 =
x02 =
2x2 + x3
: 0
x3 = x2 2x3
is asymptotically stable.
Solution. The coefficient matrix is
0
2
1
@
0
2
A=
0
1
1
1
1 A.
2
To find the eigenvalues of A we solve the equation det(A
namely
2
1
0
0
1
1
2
1
=
2
⇥
( + 2) ( + 2)2
I) = 0,
⇤
1 =0
The solutions are = 2 and + 2 = ±1, namely = 1, 3. Since
all the eigenvalues of A are negative, the origin is asymptotically stable.
10. Study the stability of the equilibrium of the system
8 0
< x1 = ax1 + 5x3
x0 =
x2 2x3
: 20
x3 =
3x3
depending on a 6= 0.
Solution. The coefficient matrix is
0
a
0
1
A=@ 0
0
0
1
5
2 A
3
This is a triangular matrix show eigenvalues are a, 1, 3. If a < 0,
all the eigenvalues are negative and hence the equilibrium is asymptotically stable. If a > 0 one eigenvalue is positive and the equilibrium is
unstable.
173
11. Show that the equilibrium of the system
8 0
< x1 = x1 + x2 + x3
x0 = x1 2x2 x3
: 20
x3 = x2 x3
is unstable.
Solution. The coefficient matrix is
0
1
1
2
A=@ 1
0
1
whose eigenvalues are the solution of
def
P ( ) = det(A
1
1
1 A
1
1
I) =
1
1
0
1
1
2
1
= 0.
1
Developing the determinant along the first column we find
P ( ) = (1
= (1
=
3
) · [( 2
)( 1
) + 1] [( 1
)(2 + )(1 + ) + (1
)+ +2
2 2 + + 5.
)
1]
Instead of finding the solutions of the third order algebraic equation
P ( ) = 0, we can argue as follows. Since P (1) = 3 > 0 and P (2) = 9,
then by the intermediate value theorem, the equation P ( ) = 0 has at
least one positive solution 2 (1, 2). This suffices tho say that the
equilibrium is unstable.
12. Find a such that the equilibrium of the system
8 0
< x1 = ax1
x0 = ax2 + x3
: 20
x3 = x2 + ax3
is asymptotically stable.
Solution. The eigenvalues of the coefficient matrix
0
1
a 0 0
A=@ 0 a 1 A
0 1 a
are
= a, a + 1, a
1. Thus for a <
1 we have asymptotic stability.
174
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
13. Study the stability of the equilibrium of the system
8 0
x = x2 + x4
>
>
< 10
x2 =
x2 + x3
0
x
=
x
>
2 + x3
>
: 30
x4 = x1 x4
Solution. Let us evaluate the eigenvalues of the coefficient matrix
0
1
0
1 0
1
B 0
1 1
0 C
C.
A=B
@ 0
1 1
0 A
1
0 0
1
Developing the determinant along the first column, one has
1
det(A
I) =
0
0
1
0
1
1
1
0
1
0
1
·
=
1
0
0
1
1
1
0
0
0
1
0
1
0
1
1
1
1
1
0
0
1
Developing the two last determinants along the third column, one finds
det(A
1
I) = ( + 1) ·
=(
2
+
So, an eigenvalue of A is
1
1
1
1
p
1
1
1
1) ·
=
1
1
1
=(
1
2
+
1)(
2
2).
2 and hence we have instability.
14. Consider the third order equation x000 + ax00 + bx0 + cx = 0 and prove
that the roots of the characteristic equation 3 + a 2 + b + c = 0
coincide with the eigenvalues of the equivalent first order system
8 0
< x1 = x2
x0 = x3
: 20
x3 =
ax3 bx2 cx1
175
Solution. The matrix of coefficients
0
0
A=@ 0
c
of the system is
1
1
0
0
1 A
b
a
The eigenvalues of A are the solutions of
1
det(A
=
[
(
I) =
a) + b]
0
c
c=
0
1
b
[
=
a
3
+a
2
+ b + c] = 0,
which is, up to the sign, the characteristic equation of x000 + ax00 + bx0 +
cx = 0.
15. Study the stability of the trivial solution x = 0 for the equations x000 +
x = 0 and x000 x = 0.
Solution. The characteristic equation of x000 + x = 0 is
One root is 1 = 1.pFactoring, we have 3 + 1 = ( + 1)(
1 ± 3i
and thus 2,3 =
. It follows that x = 0 is unstable.
2
In the case of x000 x = 0, the characteristic equation is
One root is = 1 and hence x = 0 is still unstable.
3
2
3
+ 1 = 0.
+ 1)
= 0.
16. Study the stability of the trivial solution of x000 + 5x00 + 9x0 + 5x = 0.
Solution. The characteristic equation is 3 + 5 2 + 9 + 5 = 0. By
inspection, one finds that a solution is = 1. Dividing by + 1 we
find 3 + 5 2 + 9 + 5 = ( + 1)( 2 + 4 + 5). Then the other two
solutions are = 2 ± i. Then x = 0 is asymptotically stable.
17. Prove that the trivial solution of x0000 + x000
x0
x = 0 is unstable.
Solution. One root of the characteristic equation
is = 1.
4
+
3
1=0
18. Prove that x = 0 is asymptotically stable for x0000 + 8x000 + 23x00 + 28x0 +
12 = 0.
Solution. The characteristic equation is 4 +8 3 +23 2 +28 +12 = 0.
Possible integer roots are divisors of 12. An inspection shows that the
solutions are = 1, 2, 2, 3. Thus x = 0 is asymptotically stable.
176
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
19. Prove that the equilibrium of the system
⇢ 00
x = x0 2y
y 00 =
3x + 2y 0
is unstable.
Solution. The equivalent first order system in the unknown x, u, y, v
is
8 0
x = u
>
>
< 0
u = u 2y
y0 = v
>
>
: 0
v =
3x + 2v
The equilibrium is (0, 0, 0, 0). The
0
0
B 0
A=B
@ 0
3
whose eigenvalues are the
coefficients matrix is
1
1
0 0
1
2 0 C
C
0
0 1 A
0
0 2
that solve the equation
1
def
P ( ) = det(A
0
2
0 1
0
0
3
0
I) =
0
0
1
0
2
0
0
3
2
= 0.
Developing the determinant we find
1
P( ) =
·
2
0
0
=
0
2
(1
0
1
2
)(2
)
0
0
1
=
2
6.
Then P (0) = 6 < 0 (in particular A is non singular) while P (3) =
12 > 0 . Hence P ( ) = 0 has a positive solution and thus A has a
positive eigenvalue and the equilibrium is unstable.
20. Show that the equilibrium of the system
⇢ 00
x
2y 0 = ax + 3y
00
y + 2x0 = 3x + ay
177
is unstable provided |a| < 3.
Solution. The system is equivalent to
8 0
x = u
>
>
< 0
y = v
u0 = 2v + ax + 3y
>
>
: 0
v =
2u + 3x + ay
whose coefficient matrix is
0
0
B 0
A=B
@ a
3
0
0
3
a
1
0
0
2
The eigenvalues of A solve
1
0
1 C
C .
2 A
0
0
def
P ( ) = det(A
I) =
0
a
3
3
a
1
0
0
1
2
= 0.
2
One has
0
P( ) =
=
=
2
·
3
a
2
1
2
0
+ a
3
( 2 + 4)
( 6+a )+ ( a
4
+ (4 2a) 2 + a2 9.
1
2
3
a
6) + a2
=
9
From |a| < 3 it follows that P (0) = a2 9 < 0 (in particular, A is
nonsingular because det(A) = P (0) < 0). Furthermore one has that
P ( ) ! +1 as ! +1. Then P ( ) = 0 has a positive solution.
Otherwise, one can solve the bi-quadratic equation P ( ) = 0 and check
that it has one positive real solution because a2 9 < 0. In conclusion,
A has a positive eigenvalue and hence the equilibrium is unstable.
21. Show that x0 = x
x5 has a pitchfork bifurcation.
178
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
Solution. Let f (x) = x x5 . One has f 0 (0) = and hence x = 0 is
asymptotically stable for < 0 and unstable for > 0. The equation
f (x) = 0 has nontrivial solutions if and only if > 0, given by ±x =
±( )1/4 . They form a branch such that x ! 0 as ! 0.
22. Show that x0 = x
x3
x5 has a pitchfork bifurcation.
Solution. Let f (x) = x x3 x5 . Since f 0 (0) = , x = 0 is
asymptotically stable for < 0 and unstable for > 0. For > 0
there are two more equilibria ±x that solve q
the bi-quadratic algebraic
equation x4 + x2
zero as ! 0.
23. Show that x0 = x
that k > 1.
= 0. One finds x = ±
x3
p
1+ 1+4
2
which tend to
x2k+1 has a pitchfork bifurcation provided
Solution. Let f (x) = x x3 x2k+1 . Since k > 1 (actually k > 0
suffices) we find f 0 (0) = and hence the equilibrium x = 0 is asymptotically stable for < 0 and unstable for > 0. To find nontrivial
equilibria for 6= 0, we solve for > 0 the equation x3 + x2k+1 = x,
that is x2 + x2k = . Since k > 1, this equation has exactly two
solutions ±x and x ! 0 as ! 0.
24. Show that (0, 0, 0) is unstable for the
8 0
< x1 =
x0 =
: 20
x3 =
linear system
x1
2x2
x3
and find the stable and unstable manifold.
Solution. The eigenvalues of
0
A=@
1
0
0
1
0 0
2 0 A
0 1
are 1, 2, 1. Thus (0, 0, 0) is unstable. For all p = (p1 , p1 , p3 ) the
solution x(t, p) has components given by x1 = p1 e t , x2 = p2 e 2t , x3 =
p3 et . Since limt!+1 x(t, p) = 0 if and only if p3 = 0, then the stable
manifold is the 2 dimensional linear space x3 = 0. On the other hand,
limt! 1 x(t, p) = 0 if and only if p1 = p2 = 0 and therefore the unstable
manifold is the x3 axis (x1 = x2 = 0).
179
25. Determine the stability of (0, 0) of
⇢ 0
x = x + y + y2
y 0 = 2y x2
Solution. The linearized system is
⇢ 0
x = x+y
y 0 = 2y
whose matrix has eigenvalues
stable.
1, 2. Thus (0, 0) is asymptotically
26. Show that V (x, y) = 14 (x4 + y 4 ) is a Liapunov function for the system,
⇢ 0
x = x3
y0 = y3
and deduce the stability of (0, 0).
Solution. V (x, y) > 0 and V̇ = Vx x0 + Vy y 0 = x6 y 6 < 0 for all
(x, y) 6= (0, 0). Then V is a Liapunov function and (0, 0) is asymptotically stable.
27. Show that V (x, y) = 12 (x2 + y 2 ) is a Liapunov function for the system,
⇢ 0
x = y x3
y0 = x y3
and deduce the stability of (0, 0).
Solution. V > 0 for all (x, y) 6= (0, 0). Moreover, V̇ = Vx x0 + Vy y 0 =
x(y x3 ) + y( x y 3 ) = x4 y 4 < 0 for all (x, y) 6= (0, 0). Then V
is a Liapunov function and (0, 0) is asymptotically stable. Notice that
we cannot apply the theorem on the linearized stability.
28. Show that (0, 0) is unstable for
⇢ 0
x = y + x3
y0 = x + y3
Solution. Take W (x, y) = 12 (x2 + y 2 ). One has that W > 0 for all
(x, y) 6= (0, 0) and Ẇ = xx0 + yy 0 = x(y + x3 ) + y( x + y 3 ) = x4 + y 4 >
0 for all (x, y) 6= (0, 0). Then we can apply the general instability
theorem.
180
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
29. Consider the the system
⇢
x0 = y
y 0 = x(x + a)
y
where a > 0. Show that (0, 0) is asymptotically stable.
Solution. The linearized system at (0, 0) is
⇢
x0 = y
y 0 = ax
y
whose coefficient matrix is
A=
✓
0
a
1
1
◆
.
The eigenvalues
of A are the solutions of 2 + + a = 0, that is
p
1± 1 4a
=
. If 4a > 1 the eigenvalues are complex conjugate with
2
negative real part. If 0 < 4a  1 the eigenvalues are real and negative.
Thus in both the cases (0, 0) is asymptotically stable.
30. For the same system, show that ( a, 0) is unstable.
Solution. Solving the algebraic system
⇢
0 =y
0 = ax y
we see that ( a, 0) is another equilibrium. Making the change of variable x
e = x + a the given system becomes
⇢ 0
x
e =y
y 0 = (e
x a)e
x y
The linearization at x
e = 0, y = 0 is
⇢ 0
x
e =y
y 0 = ae
x
y
whose coefficient matrix is
e=
A
✓
0
a
1
1
◆
181
p
e are e = 1± 1+4a which are real (recall that
The eigenvalues of A
2
a > 0) and with opposite sign. Then x
e = 0, y = 0 is unstable. This
obviously implies that ( a, 0) is unstable for the original x, y system.
31. Study the stability of the equilibrium of gradient system
⇢ 00
x + 4x(x2 + y 2 ) = 0
y 00 + 4y(x2 + y 2 ) = 0
Solution. Setting u = (x, y) the system is a gradient system u00 +
rF (u) with F (x, y) = (x2 +y 2 )2 . The potential F has a strict minimum
at u = 0 which is therefore stable.
32. Study the stability of the equilibrium of the equation x00 +f (x0 )+g(x) =
0 under the assumption that f (0) = g(0) = 0 and yf (y)
0 and
xg(x) > 0 for all x 6= 0.
Solution. The equation is equivalent to the system
⇢ 0
x =y
y 0 = f (y) g(x)
whose
equilibrium is (0, 0). Set V (x, y) = 21 y 2 + G(x), where G(x) =
Rx
g(s)ds. Since xG0 (x) = xg(x) > 0 for all x 6= 0, it follows that there
0
exists a neighborhood U of x = 0 such that G(x) > 0 on U \ {0}. Then
V has a local strict minimum at (0, 0). Moreover,
V̇ = Vx x0 + Vy y 0 = Vx y
= g(x)y
yf (y)
Vy [f (y) + g(x)]
yg(x) =
yf (y)  0.
Thus V is a Liapunov function and (0, 0) is stable.
33. Study the stability of the equilibrium of gradient system
⇢ 00
x + 2(x 1) + 2xy 2 = 0
y 00 + 2x2 y
=0
Solution. If u = (x, y) the system has the form u00 + rF (u) = 0
with F (u) = F (x, y) = (x 1)2 + x2 y 2 . To find the equilibria we solve
rF (u) = 0, namely
⇢
2(x 1) + 2xy 2 = 0
2x2 y
=0
182
CHAPTER 12. SOLUTIONS TO EXERCISES OF CHAPTER 12
From the second equation, either x = 0, which does not solve the first
equation, or y = 0 which yields x = 1. Thus the only equilibrium of
the system is u = (1, 0). Since F has a strict minimum at u = (1, 0),
then u is stable.
Chapter 13
Solutions to exercises of
Chapter 13
1. Show that the boundary value problem
⇢
x00 x3 = 0
x(0) = x(b) = 0
has only the trivial solution x(t) ⌘ 0.
Solution. Multiplying x00 (t) = x3 (t) by x(t) and integrating we get
R b 00
Rb 4
x
(t)x(t)dt
=
x (t)dt. Integrating by parts, taking into account
0
0
Rb
the boundary conditions x(0) = x(b) = 0, we find 0 x00 (t)x(t)dt =
R b 02
R b 02
Rb
x (t)dt. Thus
x (t)dt = 0 x4 (t)dt which implies that x(t) ⌘
0
0
0.
2. Let a < b. Prove that the boundary value problem
⇢
x00 + 4x3 = 0
x(a) = x(b) = 0
has infinitely many solutions
Solution. According to the theory we have to solve
R xc
b a
(c) = p ,
2k
dx
and xc = c1/4 . Setting x = c1/4 z we find
4
c
x
p
p
p
dx = c1/4 dz and c x4 = c(1 z 4 ) = c1/2 1 z 4 . Then
Z 1
Z 1
c1/4 dz
dz
1/4
p
p
(c) =
=c
.
1/2
4
1 z
1 z4
0 c
0
where
(c) =
0
p
183
184
CHAPTER 13. SOLUTIONS TO EXERCISES OF CHAPTER 13
Setting A =
equation A c
R1
0
1/4
dz
> 0 we find (c) = A c 1/4 . Clearly the
4
1 z
b a
= p
has infinitely many solutions
2k
" p #4
A 2k
ck =
,
k = 1, 2, ....
b a
p
3. Show that the boundary value problem
⇢
x00 + 6x5 = 0
x(0) = x(b) = 0
has infinitely many solutions.
Solution. Here
(c) =
Z
xc
0
p
dx
,
c x6
where xc = c1/6 .
p
p
Setting x = c1/6 z we find dx = c1/6 dz and c x6 = c1/2 1 z 6 .
Thus
Z 1
dz
1/3
p
(c) = Ac
, where A =
> 0.
1 z6
0
b
b
Then the equation (c) = p , namely Ac 1/3 = p , has infinitely
2k
2k
many solutions
" p #3
A 2k
ck =
,
k = 1, 2, ....
b
4. Show that for all k
0 the boundary value problem
⇢
x00 + (2p + 2)x2p+1 = 0
x(0) = x(b) = 0
has infinitely many solutions.
Solution. Here
(c) =
Z
xc
0
p
dx
,
c x2p+2
where xc = c1/(2p+2) .
185
p
1/(2p+2)
1/(2p+2)
Setting
x
=
c
z
we
find
dx
=
c
dz
and
c x2p+2 =
p
1/2
2p+2
c
1 z
. Thus
Z 1
dz
p/(2p+2)
p
(c) = Ac
, where A =
.
1 z 2p+2
0
b
b
(c) = p , namely Ac p/(2p+2) = p , has in2k
2k
finitely many solutions
" p #(2p+2)/p
A 2k
ck =
,
k = 1, 2, ....
b
Then the equation
5. Show that for
< ⇡ the boundary value problem
⇢
x00 + x x3 = 0
x(0) = x(1) = 0
has only the trivial solution.
Solution. It suffices to notice that ⇡ is the first eigenvalue of the
linearized problem x00 + x = 0, x(0) = x(1) = 0.
6. Show that the following boundary value problems
⇢
⇢
x00 + 4x3 = 0
x00 + 4x3 = 0
(a)
(b)
x(0) = 0, x(b) = 1
x0 (0) = 0, x(b) = 0
have a positive solution.
1 2
y +
2
x4 = c. For the problem (a) we take the arc ⇤c in the first quadrant
between the straight lines x = 0 and x = 1, see fig. 13.1. This arc
corresponds to a positive solution of (a) provided b = (c) where
Z 1
Z 1
dx
1
dx
p
(c) =
=p
y
c x4
2 0
0
Solution. In the phase plane consider the curve ⇤c of equation
Setting x = c1/4 z we find
Z c 1/4
Z
1
c1/4 dz
c 1/4 c
p
(c) = p
= p
c1/2 1 z 4
2 0
2 0
1/4
p
dz
.
1 z4
186
CHAPTER 13. SOLUTIONS TO EXERCISES OF CHAPTER 13
y
x
1
Figure 13.1: The curve ⇤c in black, the arc ⇤c in red
Rc
dz
one has
1 z4
Z +1
dz
p
lim g(c) =
> 0,
c!0+
1 z4
0
Setting g(c) =
Then
1/4
0
p
lim g(c) = 0.
c!+1
c 1/4
(c) = p g(c) satisfies
2
lim
c!0+
(c) = +1,
lim
c!+1
(c) = 0.
Since (c) is continuous, it follows that the equation b =
solution, yielding a positive solution of problem (a).
(c) has a
b c ⇢ ⇤c contained in
In the case of problem (b) we consider the arc ⇤
the fourth quadrant, see fig. 13.2.
Here we have to solve
c 1/4
b= p
2
Z
1
0
p
dz
1 z4
187
y
1/4
c
x
b c (in red).
Figure 13.2: Plot of 12 y 2 + x4 = c (in black) and of the arc ⇤
which has the solution

A
c= p
b 2
4
,
A=
Z
1
0
p
dz
.
1 z4
7. Prove that the preceding problems (a) and (b) have infinitely many
solutions.
Solution. According to the previous exercise, it suffices to solve
Z 1/4
kc 1/4 c
dz
p
(a)
b= p
,
1 z4
2 0
respectively
Z
kc 1/4 1
dz
p
(b)
b= p
.
1 z4
2 0
Both these equations have infinitely many solutions ck , k = 1, 2, ..
8. Find b > 0 such that the boundary value problem
⇢
x00 + 4x3 = 0
x(0) = 0, x(b) = 1, x0 (b) = 0
188
CHAPTER 13. SOLUTIONS TO EXERCISES OF CHAPTER 13
has a positive solution.
Solution. In the phase plane consider the curve ⇤ of equation 12 y 2 +
e ⇢ ⇤ contained in the first quadrant, see fig.13.3,
x4 = 1. The arc ⇤
y
√2
1
x
e (in red).
Figure 13.3: Plot of 12 y 2 + x4 = 1 (in black) and of the arc ⇤
carries a positive solution of x00 + 4x3 = 0 such that x(0) = 0, x(t⇤ ) = 1
and x0 (t⇤ ) = 0, where
⇤
t =
Z
1
0
dx
1
=p
y
2
Z
1
0
p
dx
.
1 x4
To find a solution of our problem we impose that t⇤ = b. Thus
1
b= p
2
Z
1
0
p
dx
.
1 x4
Notice that the previous integral is (positive and) finite.
9. Find the Green function of L[x] = x00 on [0, 1] and solve the boundary
value problem x00 = 1, x(0) = x(1) = 0.
189
b
-1
Figure 13.4: Plot of f (x) =
1
p1 p 1
.
2 1 x4
x
The shadowed area equals b.
Solution. Take '(t) = t and (t) = 1
dent and W = 1. Hence
⇢
t(1 s),
G(t, s) =
s(1 t),
t. They are linearly indepenif t 2 [0, s],
if t 2 [s, 1].
Using the Green function, the solution of the boundary value problem
L[x] + h(t) = 0, x(0) = x(1) = 0 is given by
Z 1
x(t) =
G(t, s)h(s)ds.
0
R1
Rt
R1
To evaluate the integral we split the integral 0 ds into 0 ds + t ds.
Since G(t, s) = (1 t)s for 0  s  t, while G(t, s) = t(1 s) for
t  s  1 and since h(s) = 1 we find
Z t
Z 1
x(t) = (1 t) ( s)ds + t
(s 1)ds
0
t
✓
◆
1 2 1 3
1
1 2
=
t + t +t
+t
t
2
3
2
2
1
1
= t2
t.
2
2
190
CHAPTER 13. SOLUTIONS TO EXERCISES OF CHAPTER 13
10. Find the Green function of L[x] = x00 on [ 1, 1].
Solution. Take '(t) = t + 1 and (t) = 1 t. They are linearly
independent and W = 1. Hence
⇢
(t + 1)(1 s),
if t 2 [ 1, s]
G(t, s) =
(s + 1)(1 t),
if t 2 [s, 1]
11. Find the Green function of L[x] = x00
k 2 x on [0, 1], where k 6= 0.
Solution. The general solution of x00 k 2 x = 0 is x = c1 ekt + c2 e kt .
If we impose x(0) = 0, x0 (0) = 1 we find c1 + c2 = 0 and kc1 kc2 = 1,
1
c2
that is c1 = c2 =
. If x(1) = 0, x0 (1) = 1 we find c1 e +
=0
2k
e
k
and c1 ke c2 = 1. Solving the first equation we find c2 = c1 e2 .
e
Substituting into the second,
c1 ke + c1 e2
namely 2c1 ke =
k
=
e
1,
1 and thus
c1 =
1
,
2ke
c2 =
e
.
2k
Therefore, we can take
'(t) =
1 kt
(e
2k
e
kt
)=
1
sinh(kt)
k
and
1 k(t
(e
2k
Since the wronskian is
(t) =
W =
'0 (0) (0) =
1)
e
k(t 1)
(0) =
)=
1
sinh[k(t
k
1
sinh( k) =
k
1)].
1
sinh(k),
k
we infer that
G(t, s) =
8
1
>
· sinh(kt) · sinh[k(s
>
>
< k sinh(k)
>
>
>
:
1
· sinh(ks) · sinh[k(t
k sinh(k)
1)],
if t 2 [0, s],
1)],
if t 2 [s, 1].
191
12. Show that x00 = 1
that 0  x(t)  1.
x2 , x(a) = x(b) = 0 has a solution x(t) such
x
Solution. It suffices to remark that v ⌘ 0 is a sub solution and w ⌘ 1
is a super solution.
13. Show that x00 + x = e x , x(a) = x(b) = 0 has a solution such that
0  x(t)  1.
Solution. Writing the equation as x00 = e x x we deduce that
v ⌘ 0 is a sub solution and w ⌘ 1 is a super solution.
14. Let g(x) be continuous and such that g(0) > 0, g(1) < 1. Show that
x00 + x = g(x), x(a) = x(b) = 0 has a solution.
Solution. Writing the equation as x00 = g(x) x we infer that v ⌘ 0
is a sub solution and w ⌘ 1 is a super solution.
15. Show that
x00 + x = e
x2
, x(a) = x(b) = 0 has a positive solution.
2
Solution. Since 0 < e x  1, then v = 0 is a sub solution and
w = 1 is a super solution. Then there is a solution x(t) such that 0 
x(t)  1. Moreover, if x(⌧ ) = 0 for some ⌧ 2]a, b[ then ⌧ is a minimum,
2
whilst from the equation one infers x00 (⌧ ) = e x (⌧ ) = 1 < 0, a
contradiction. Thus x(t) > 0 for all t 2]a, b[.
16. Let g(x) be continuous and such that 0 < g(x)  M for all x. Show
that x00 = g(x) x, x(a) = x(b) = 0 has a positive solution.
Solution. v = 0 is a sub solution and w = M is a super solution. Then
there is a solution x(t) such that 0  x(t)  M . If x(⌧ ) = 0 for some
⌧ 2]a, b[ then ⌧ is a minimum, whilst x00 (⌧ ) = g(x(⌧ )) = g(0) < 0.
Thus x(t) is a positive solution.
17. Show that
solution.
x00 = (1 + x2 )
1/2
Solution. One has 0 < (1 + x2 )
18. Show that
solution.
x, x(a) = x(b) = 0 has a positive
1/2
 1.
x00 + x = min{ex , 1}, x(a) = x(b) = 0 has a positive
Solution. One has 0 < min{ex , 1}  1.
192
CHAPTER 13. SOLUTIONS TO EXERCISES OF CHAPTER 13
19. Prove that
x00 = 2x
x2 , x(0) = x(⇡) = 0 has a positive solution.
Solution. The equation has the form x00 = 2x g(x) with g(x) = x2 .
Then
g(x)
lim
= lim x = +1.
x!+1 x
x!+1
Moreover, 2 is greater than the first eigenvalue 1 = 1 of x00 =
x, x(0) = x(⇡) = 0. Then there exists a positive solution.
20. Show that if b > ⇡, the problem
a positive solution.
x00 = arctan x, x(0) = x(b) = 0 has
Solution. Write arctan x = x g(x) with g(x) = arctan x
apply Theorem 13.4.1 with = 1.
x and