Name:
November 8, 2011
Worksheet # 2: Higher Order Linear ODEs (SOLUTIONS)
1. A set of n-functions f1 , f2 , ..., fn are linearly independent on an interval I if the only way that
c1 f1 (t) + c2 f2 (t) + ... + cn fn (t) = 0
holds for all t in I is if c1 = c2 = ... = cn = 0. Otherwise, they are linearly dependent.
(a) Prove that y1 (t) = t2 , y2 (t) = 3t2 + t + 1, y3 (t) = t + 1, y4 = sin t are linearly dependent on
(−∞, ∞) by finding a combination with c1 , ..., c4 not all zero so that
c1 y1 (t) + c2 y2 (t) + c3 y3 (t) + c4 y(t) = 0
for all t.
Solution. Set c1 = −3, c2 = 1, c3 = −1 and c4 = 0, since
−3(t2 ) + (3t2 + t + 1) − (t + 1) + 0(sin t) = 0
for all t. Note there are other possible choices.
(b) There is a Wronskian test for verifying that a set of
twice-differentiable functions, the Wronskian is
y1 (t)
W (y1 , y2 , y3 )(t) = y10 (t)
y100 (t)
functions is linearly independent. For 3
y2 (t)
y20 (t)
y200 (t)
y3 (t) y30 (t) y300 (t)
(The Wronskian for 4 or more functions is defined in a similar way (see page 221) but requires
computation of determinant larger than 3 × 3). If W (y1 , y2 , y3 )(t) 6= 0 for some point t in
I, then the functions must be linearly independent. If, on the other hand, the functions are
linearly dependent then the Wronskian must be 0 for all t in I.
i. Use the Wronskian to verify that y1 = x, y2 = x2 and y3 = x3 are linearly independent
on (−∞, ∞).
Solution.
x x2 x3 W (y1 , y2 , y3 )(x) = 1 2x 3x2 = x(12x2 − 6x2 ) − x2 (6x) + x3 (2) = 2x3 .
0 2
6x Since this expression is not zero for example at x = 1 the functions are linearly independent.
ii. The functions y1 (t) = t2 + t, y2 (t) = 2t2 − t, and y3 (t) = 2t are linearly dependent on
(−∞, ∞). Find c1 , c2 , c3 not all zero so that
c1 y1 (t) + c2 y2 (t) + c3 y3 (t) = 0
holds for all t. Verify that the Wronskian is 0 for all t.
Solution. Let c1 = −2, c2 = 1, and c3 = 3/2. We have
3
−2(t2 + t) + 1(2t2 − t) + (2t) = −2t2 − 2t + 2t2 − t + 3t = 0.
2
Therefore, the functions are linearly dependent. The Wronskian must be zero. We check
that
2
t + t 2t2 − t 2t
W (y1 , y2 , y3 )(t) = 2t + 1 4t − 1 2 2
4
0
= (t2 + t)(−8) − (2t2 − t)(−4) + 2t(4(2t + 1) − 2(4t − 1))
= −8t2 − 8t + 8t2 − 4t + 2t(8t + 4 − 8t + 2)
= −12t + 12t = 0
2. Consider the homogeneous third order differential equation
L[y] = y (3) + p(t)y 00 + q(t)y 0 + r(t)y = 0
Assume that p, q, r are all continuous functions on an open interval I. The general solution to this
equation is
y = c1 y1 + c2 y2 + c3 y3
where y1 , y2 , and y3 are three times differentiable, linearly independent functions which are also
solutions to the ODE. The solutions y1 , y2 , y3 form a fundamental set. There is again a Wronskian
test for linearly independence of solutions: y1 , y2 , y3 are linearly independent solutions to the ODE
if and only if
y1 (t) y2 (t) y3 (t) 0
W (y1 , y2 , y3 )(t) = y1 (t) y20 (t) y30 (t) 6= 0
y100 (t) y200 (t) y300 (t)
for some point t in I.
(a) Let p(t) = −6, q(t) = 11, and r(t) = −6. Verify that y1 (t) = et , y2 (t) = e2t , y3 (t) = e3t
are linearly independent solutions to the ODE. Construct the general solution and find the
constants c1 , c2 , c3 so that the initial conditions
y(0) = 1,
y 0 (0) = −1,
y 00 (0) = 1
hold.
Solution. The differential equation is
L[y] = y 000 − 6y 00 + 11y 0 − 6y = 0
Note that
L[ert ] = (r3 − 6r2 + 11r − 6)ert
so that if y = ert is a solution r must be a root of the cubic equation
r3 − 6r2 + 11r − 6 = 0
This equation factors as
(r − 1)(r − 2)(r − 3) = 0
There are three, distinct real roots r = 1, 2, 3. Each of these gives a solution to the differential
equation: y1 (t) = et , y2 (t) = e2t , and y3 (t) = e3t . We must verify that these are linearly
independent. Compute the Wronskian:
t
2t
3t e
t e 2t e 3t 3e W (y1 , y2 , y3 )(t) = e 2e
et 4e2t 9e3t = et (18e5t − 12e5t ) − e2t (9e4t − 3e4t ) + e3t (4e3t − 2e3t )
= 6e6t − 6e6t + 2e6t
= 2e6t 6= 0
So the solutions are linearly independent.
The general solution is
y(t) = c1 et + c2 e2t + c3 e3t
The first and second derivatives of this solution are
y 0 (t) = c1 et + 2c2 e2t + 3c3 e3t
y 00 (t) = c1 et + 4c2 e2t + 9c3 e3t
To satisfy the initial conditions at t = 0 we need:
1 = c1 + c2 + c3
−1 = c1 + 2c2 + 3c3
1 = c1 + 4c2 + 9c3
There are many ways to solve this system of algebraic equations. One easy way is to subtract
the first equation from the third:
0 = 3c2 + 8c3
So
8
c2 = − c3 .
3
We can then subtract the second equation from the first to get
2 = −c2 − 2c3
Substituting the c2 = − 38 c3 :
2
c3
3
2=
⇒
c3 = 3
Then
c2 = −8
and plugging into the first equation
1 = c1 − 8 + 3
⇒ c1 = 6.
Therefore, the unique solution satisfying these initial conditions is
y = 6et − 8e2t + 3e3t
4.1: #11 Verify that 1, cos t, sin t are solutions to
y 000 + y 0 = 0
and compute their Wronskian.
Solution. By substituting these functions into the equation, it is clear that each is a solution.
We compute the Wronskian
1 cos t
sin t 0 − sin t cos t = 1(sin t2 + cos t2 ) = 1.
0 − cos t − sin t
4.1: # 15 Verify that 1, x, x3 are solutions to
xy 000 − xy 00 = 0
and compute their Wronskian.
Solution. By substituting these functions into the equation, it is clear that each is a solution.
For example
x(x3 )000 − x(x3 )00 = 3x(x2 )00 − 3x(x2 )0 = 6x(x)0 − 6x = 0.
We compute the Wronskian
1
0
0
x x3 1 3x2 = 6x.
0 6x 3. Consider the constant coefficient homogeneous, 3rd order ODE
ay 000 + by 00 + cy 0 + dy = 0.
(a) Substitute y = ert into the ODE to obtain a characteristic equation. You can always factor
this equation into the form
a(r − r1 )(r − r2 )(r − r3 ) = 0.
List all possible cases for the roots of r. For example, can this equation have three complex
roots? One complex and two real? Repeated roots?
Answer. The important fact is that complex roots of polynomial equations (with real coefficients) can only occur in complex conjugate pairs. So you cannot have three complex roots or
one complex root. Since the roots are conjugates, complex roots in this case can’t be repeated.
The cases are
I.
II.
III.
IV.
Three real and distinct roots.
Three real roots, but only two distinct. That is r1 , r2 , r3 real with, say r2 = r3 .
Three real roots, all repeated.
One real root and a complex conjugate pair of roots.
(b) If the roots are all real and different, prove that y1 = er1 t , y2 = er2 t , y3 = er3 t form a
fundamental set.
Solution. Use the operator notation
L[y] = ay 000 + by 00 + cy 0 + dy.
If y = ert , then it is clear that
L[y] = (ar3 + br2 + cr + d)ert
and so L[y] = 0 if and only if r is one of r1 , r2 , r3 . Thus y1 , y2 , y3 are solutions. The Wronskian
is
rt
e1
er2 t
er3 t rt
r1 e 1 r2 er2 t r3 er3 t = (r2 r32 − r22 r3 )e(r1 +r2 +r3 )t − (r1 r32 − r12 r3 )e(r1 +r2 +r3 )t
2 rt
r1 e 1 r22 er2 t r32 er3 t + (r1 r22 − r12 r2 )e(r1 +r2 +r3 )t
= (r2 r32 − r22 r3 − r1 r32 − r12 r3 + r1 r22 − r12 r2 )e(r1 +r2 +r3 )t
= (r2 − r1 )(r3 − r1 )(r3 − r2 )e(r1 +r2 +r3 )t
This expression is clearly not zero since all of r1 , r2 and r3 are distinct. Therefore, the
solutions are linearly independent.
(c) If the equation has one repeated real root r, prove that y1 = ert , y2 = tert , y3 = t2 ert form a
fundamental set.
Solution. You can verify exactly as above that y1 (t) = ert is a solution. Compute
L[y2 ] = a(tert )000 + b(tert )00 + c(tert )0 + dtert
= a(ert + rtetr )00 + b(ert + rtetr )0 + c(ert + rtetr ) + dtert
= a(rert + retr + r2 tert )0 + b(rert + retr + r2 tert )
+ c(ert + rtert ) + dtert
= a(3r2 ert + r3 tert ) + b(2r + r2 t)ert
+ c(1 + rt)ert + dtert
= (ar3 + br2 + cr + d)tert + (3ar2 + 2br + c)ert
Now clearly
ar3 + br2 + cr + d = 0.
I claim the second term is also zero. To see why, notice that
ax3 + bx2 + cx + d = a(x − r)3
Expanding the right hand side we see that
ax3 + bx2 + cx + d = a x3 − 3 a r x2 + 3 a r2 x − a r3 .
Matching coefficients, we have that
b = −3ar,
c = 3ar2 ,
d = −ar3 .
Therefore
3ar2 + 2br + c = 0
and it follows that
L[y2 ] = 0.
The same trick will work to show that
L[y3 ] = 0
and y1 , y2 , y3 are all solutions.
We compute the Wronskian and carefully simplify to obtain
rt
e
tert
t2 ert
rt
rt
2 rt
re
= 2e3rt 6= 0
(1
+
rt)e
(2t
+
rt
)e
2 rt
r e
r(2 + rt))ert (2 + 4rt + r2 t2 )ert So the solutions y1 , y2 , y3 are linearly independent on (−∞, ∞) and hence form a fundamental
set for the ODE.
(d) Complex roots always occur in conjugate pairs λ ± µi. A complex conjugate pair gives rise to
solutions y1 (t) = eλt cos µt and y2 (t) = eλt sin µt. What is the fundamental solution set in this
case? Be sure to verify that the solutions in your fundamental set are linearly independent.
Solution. The characteristic equation has one real root, call it r1 , and a complex conjugate
pair of roots r2,3 = µ ± λt.
The fundamental set in this case is
y1 (t) = ert ,
y2 (t) = eλt cos µt, y3 (t) = eλt sin µt.
It is clear that y1 (t) is a solution by the arguments of the previous problem(s). You do not
have to check that y2 (t), y3 (t) are solutions, since it is a given in the statement of the problem.
If you want to see why they are though, substitute y2 (t) into the equation and simplify to
obtain:
(
( 3 (
) )
L[y2 ] =
a µ + −3 a λ2 − 2 b λ − c µ sin (µ t)
)
(
)
+ (−3 a λ − b) µ2 + a λ3 + b λ2 + c λ + d cos (µ t) eλ t
On the other hand, you can plug r2 back in to the characteristic equation and collect real and
imaginary parts to see that
0 = ar23 + br22 + cr2 + d
(
)
= (−3 a λ − b) µ2 + a λ3 + b λ2 + c λ + d
)
((
)
+ 3 a λ2 + 2 b λ + c µ − a µ3 i.
For this to hold, it must follow that
(
)
3 a λ2 + 2 b λ + c µ − a µ3 = 0
(−3 a λ − b) µ2 + a λ3 + b λ2 + c λ + d = 0
and y2 must be solution. A similar argument works for y3 .
The Wronskian to expand is:
rt
e
cos (µ t) et λ
sin (µ t) et λ
rt
tλ
tλ
re
(cos
(µ
t)
λ
−
µ
sin
(µ
t))
e
(sin
(µ
t)
λ
+
µ
cos
(µ
t))
e
2 rt (
)
(
)
r e
cos (µ t) λ2 − 2 µ sin (µ t) λ − µ2 cos (µ t) et λ
sin (µ t) λ2 + 2 µ cos (µ t) λ − µ2 sin (µ t) et λ If you compute things correctly, you should find that
W (y1 , y2 , y3 )(t) = µ(λ2 − 2rλ + r2 + µ2 )e2tλ+rt
Can this quantity ever be 0? Well, the exponential part is never 0 and µ 6= 0, since r2 , r3
must have an imaginary part. This leaves the possibility that
λ2 − 2 r λ + r2 + µ2 = 0.
But this is a quadratic equation in r and if you solve it, you will find that the roots are
precisely
r = µ ± λi.
This cannot occur since by assumption r must be the real root.
We therefore may conclude that
W (y1 , y2 , y3 )(t) 6= 0
and so the solutions are linearly independent on (−∞, ∞). Therefore, y1 , y2 , y3 form a fundamental set.
(e) Now provide the fundamental solution sets for all other cases you described in part (a).
Solutions. The only remaining case is (II). Here we have the root r1 and the twice repeated
root r2 and these lead to the fundamental set y1 (t) = er1 t , y2 (t) = ter2 t , and y3 (t) = ter3 t .
You can verify that this is a fundamental set as we did above but I do not require you to do
this.
(f) 4.2: # 29: Solve
y 000 + y 0 = 0
y(0) = 0
y 0 (0) = 1
y 00 (0) = 2
Solution. The characteristic equation is
r3 + r = r(r2 + 1) = 0.
The roots of this equation are
r1 = 0
r2,3 = ±i.
Thus the general solution is
y(t) = c1 + c2 cos t + c3 sin t.
Compute y 0 (t) and y 00 (t):
y 0 (t) = −c2 sin t + c3 cos t
y 00 (t) = −c2 cos t − c3 sin t
Using the initial conditions, we obtain the equations
c1 + c2 = 0
−c3 = 1
−c2 = 2
So c1 = 2, c2 = −2, and c3 = −1
The solution is
y(t) = 2 − 2 cos t − sin t
(g) Problem 4.2 #32 : Solve
y 000 − y 00 + y 0 − y = 0
y(0) = 2
y 0 (0) = −1
y 00 (0) = −2
Solution. The characteristic equation is
r3 − r2 + r − 1 = 0.
There are a number of ways you can find the roots of simple polynomial equations like this.
One method is to guess a root, e.g. r = 1, verify that its a root, then do polynomial division
to break the polynomial into a linear term and a quadratic term. Here r = 1 is a root and
doing polynomial division, we factor this equation as
(r − 1)(r2 + 1) = 0.
So the roots are r1 = 1 and r2,3 = ±i. The general solution is thus
y(t) = c1 et + c2 cos t + c3 sin t.
The derivatives are
y 0 (t) = c1 et − c2 sin t + c3 cos t
y 00 (t) = c1 et − c2 cos t − c3 sin t.
Using the initial conditions, we obtain the coefficient equations
c1 + c2 = 2
c1 + c3 = −1
c1 − c2 = −2
If you add the first and last equation, you obtain c1 = 0 so that c2 = 2 by the first equation
and c3 = −1 by the second. The solution is
y(t) = 2 cos t − sin t
4. Consider the the constant coefficient homogeneous n-th order ODE:
a0 y (n) + a1 y (n−1) + ... + an−1 y 0 + an y = 0.
By substituting in y = ert you will obtain a characteristic equation
a0 rn + a1 rn−1 + ... + an−1 r + an = a0 (r − r1 )(r − r2 ) · · · (r − rn ) = 0.
There are many cases for the roots here and it can be extraordinarily difficult to factor higher order
polynomial equations1 . You should use software to find roots for higher order cases. By finding
all the roots you can generate a fundamental set of solutions exactly as we did in the second order
and third order cases. Combinations of distinct real roots and complex roots work the same way
as in the previous problem. For higher order equations you can have many repeated roots and also
complex repeated roots. Consult the text page 228 to see how to handle these scenarios. Use these
ideas to solve the following problems from the text.
(a) 4.2: # 19: Find the general solution to
y (5) − 3y (4) + 3y 000 − 3y 00 + 2y 0 = 0.
Solution. The characteristic equation is
r5 − 3r4 + 3r3 − 3r2 + 2r = r(r4 − 3r3 + 3r2 − 3r + 2) = 0
It is clear that r = 0 is one root. You could use Maxima to further factor the polynomial or
guess roots and do polynomial division. Obtain
r(r − 1)(r − 2)(r2 + 1) = 0
so that the characteristic equation has roots r = 0, 1, 2, ±i. The general solution is
y(t) = c1 + c2 et + c3 e2t + c4 cos t + c5 sin t.
(b) 4.2: # 21: Find the general solution of
y (8) + 8y (4) + 16y = 0.
Solutions. The characteristic equation is
r8 + 8r4 + 16 = 0
Again you might use software to factor this but I indicate a different approach. Let u = r4 .
But you may also be able factor this by hand:
(
)2
r8 + 8r4 + 16 = (r4 + 4)2 = (r2 − 2r + 2)(r2 + 2r + 2) = 0
The roots of the first quadratic term are r1,2 = 1 ± i and the roots of the second are r3,4 =
−1 ± i. Since the quadratic factors are squared, each of these roots is repeated. This gives
the general solution
y(t) = et (c1 cos t + c2 sin t) + e−t (c3 cos t + c4 sin t)
+ et t (c5 cos t + c6 sin t) + e−t t (c7 cos t + c8 sin t)
1 There are formulas analogous to the quadratic formula which find all the roots of cubic and quartic polynomials. They
are very complicated and it is worth a Google search to see just how involved the quartic formula is. An amazing fact is
that it is impossible to write down general closed formulas for the roots of 5th and higher degree polynomials. It is not
just that none have been found but that it cannot be done!
(c) 4.2: # 30: Solve
y (4) + y = 0
y 0 (0) = 0
y(0) = 0
y 00 (0) = −1
y 000 (0) = 0
Solution. The characteristic equation is
r4 + 1 = 0
The roots of this equation are
1
1
r1,2 = √ ± √ i
2
2
1
1
r3,4 = − √ ± √ i
2
2
The general solution is
)
(
))
)
(
))
(
(
(
(
−t
1
1
1
1
√t
√
2
2
√
√
√
√
y(t) = e
t + c2 sin
t
+e
t + c4 sin
t
.
c1 cos
c3 cos
2
2
2
2
So
y(0) = c1 + c3 = 0
Compute the first through third derivative of y and evaluate at t = 0. You can do this by
hand or with a computer and use the initial conditions to obtain the equations
c1 + c3 = 0
c1 + c2 − c3 + c4 = 0
c2 − c + 4 = −1
−c1 + c2 + c3 + c4 = 0
One way to solve this system is to add the second and fourth equations to get
2c2 + 2c4 = 0 ⇒ c2 + c4 = 0
Adding the third equation to this, you’ll find that c2 =
equation is now
−c1 + c3 = 0
−1
2
and then that c4 = 12 . The fourth
and using the first equation it follows that c1 = c3 = 0.
So the solution is
( √
)
−1
1
(
)
√
t
t
e 2
e 2
t
y(t) =
−
sin √
.
2
2
2