Review Session for Final Exam
December 11, 2015
All Problems in Final Exam in 2014 Fall
1. Solve the initial value problem
(3xy + y 2 ) + (x2 + xy)y 0 = 0,
y(1) = 2.
(a) (3 points) Verify that µ(x) = x is an integrating factor, that is, x(3xy + y 2 )dx + x(x2 + xy)dy = 0 is
exact.
Answer.
Let M = x(3xy + y 2 ) = 3x2 y + xy 2 and N = x(x2 + xy) = x3 + x2 y, then
My = 3x2 + 2xy,
and Nx = 3x2 + 2xy.
Then My = Nx . So x(3xy + y 2 )dx + x(x2 + xy)dy = 0 is exact.
Remark 1. The differential equation M (x, y) + N (x, y)y 0 = 0 is exact if
My (x, y) = Nx (x, y).
(b) (5 points) Solve the initial value problem.
Answer.
Since M dx + N dy = 0 is exact. Then there exists some φ(x, y) such that
φx = M = 3x2 y + xy 2 ,
and φy = N = x3 + x2 y.
Since φx = 3x2 y + xy 2 , then
Z
φ(x, y) =
Since φy = x3 + x2 y, then
1
(3x2 y + xy 2 ) dx + g(y) = x3 y + x2 y 2 + g(y).
2
x3 + x2 y = φy = x3 + x2 y + g 0 (y),
which implies that g 0 (y) = 0, so g(y) = constant. Hence
1
x3 y + x2 y 2 = C.
2
Since y(1) = 2, then 2 + 2 = C, that is, C = 4. Therefore, the solution is:
1
x3 y + x2 y 2 = 4.
2
Remark 2. If the differential equation M (x, y) + N (x, y)y 0 = 0 is exact, that is, My (x, y) = Nx (x, y),
then there exists a function φ(x, y) such that
• φx (x, y) = M (x, y).
• φy (x, y) = N (x, y).
• The solution to M (x, y) + N (x, y)y 0 = 0 is given by φ(x, y) = C .
To find φ(x, y):
Step I: Since φx (x, y) = M (x, y), for any fixed y, we integrate with respect to x, we get
Z
φ(x, y) = M (x, y) dx + f (y), for some f (y).
Step II: Now we only
Z need to compute f (y): Take the partial derivative with respect to y on both sides of
φ(x, y) = M (x, y) dx + f (y), we get
Z
φy (x, y) =
My (x, y) dx + f 0 (y).
Since φy (x, y) = N (x, y), so we get f 0 (y) = N (x, y) −
Z
My (x, y) dx ( this is a function of y), which
implies that
f (y) =
Z Z
N (x, y) − My (x, y) dx dy.
In summary, we get
0
For exact equation: M (x, y) + N (x, y) y = 0 =⇒
Z
Z Z
M (x, y) dx +
N (x, y) − My (x, y) dx dy = C .
2. A second order chemical reaction can be modeled by the equation
dy
= α(y − p)(y − q),
dx
where α, p and q are constants
(a) (4 points) Assume that α > 0 and p > q > 0. Find equilibrium points and classify stabilities of these
equilibrium points.
Answer.
Let’s solve α(y − p)(y − q) = 0, since α > 0 and p > q > 0, then y = p or y = q. So we have
two equilibrium points:
y = p, and y = q.
The phase diagram is:
So y = p is unstable, y = q is stable.
Remark 3. If f (a) = 0, y(x) ≡ a is called an equilibrium solution, and a is called a critical point of
y 0 = f (y).
Remark 4. To draw the phase diagram of y 0 = f (y):
1) Find all critical points of y 0 = f (y), that is, find all zeros of f (y) = 0.
2) Mark all critical points on a vertical line.
3) For any two neighboring critical points a and b, choose any point c which is between a and b, compute
f (c):
– If f (c) > 0, draw an up arrow “↑” between a and b.
– If f (c) < 0, draw a down arrow “↓” between a and b.
Remark 5. Let a be a critical point of y 0 = f (y), that is, f (a) = 0. Then
I. We say a is a sink if any solution with initial condition close to a is asymptotic to a as x increases.
In this case, a sink is stable.
Figure 1: Sink(stable)
II. We say a is a source if all solutions start close to a tend toward y as x decreases, and tend away
from a as x increases. In this case, a source is unstable.
Figure 2: Source(unstable)
III. We say a is a node if the equilibrium point a is neither a source nor a sink. In this case, a node is
unstable.
Figure 3: Node(unstable)
(b) (4 points) Assume that α = 1, p = 0, q = 1 and y(0) = −1, solve the initial value problem and determine
the limiting value of y(x) as x → ∞.
Answer.
Since α = 1, p = 0 and q = 1, then our differential equation becomes
dy
= y(y − 1).
dx
So
dy
= dx, that is,
y(y − 1)
1
1
−
y−1 y
dy = dx. Then
y − 1
= x + C.
ln y Since y(0) = −1 < 0, then y(x) < 0 for all x ≥ 0, then
y−1
= x + C.
ln
y
Then
y(x) =
1
.
1 − ex+C
Since y(0) = −1, then C = ln 2. So solution is y(x) =
1
, which implies that lim y(x) = 0.
x→∞
1 − 2ex
Remark 6. A separable differential equation has the form of:
y0 =
dy
= f (x)g(y).
dx
There are two cases:
Case I: If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
Case II: If g(y) 6= 0, then moving g(y) to the left hand side, we get
1
dy
·
= f (x).
g(y) dx
Integrate both sides with respect to x, then
Z
Z
1
dy
·
dx = f (x) dx.
g(y) dx
By the change of variables, then
Z
One can just think the cancellation in
1
dy =
g(y)
Z
f (x) dx.
dy
dy
= dy.
dx, that is,
dx
dx
dx
In summary, we have


 1) y(x) ≡ a for some constant a such that g(a) = 0
Z
Z
0
y = f (x)g(y) =⇒
.
1

dy = f (x) dx.
 2)
g(y)
(c) (4 points) Assume that α = 1, p = q = 0 and y(0) = 1. Use Euler’s method to approximate y(2) with
the step size h = 1.
Answer.
Since α = 1, p = q = 0 and y(0) = 1, then our differential equation becomes
dy
= y2,
dx
and y(0) = 1.
Recall the linear approximation, we get
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
Let yk = y(xk ), put x = xk and ∆x = 1 in the above formula, so we have
yk+1 = yk + yk2 · 1 = yk + yk2 .
Let f (y) = y 2 , x0 = 1 and y0 = y(x0 ) = y(0) = 1, then
k
0
1
2
xk
0
1
2
yk
1
y1 = y0 + y02 = 1 + 12 = 2
y2 = y1 + y12 = 2 + 22 = 6
Remark 7. Consider the initial value problem:
y 0 = f (x, y),
y(x0 ) = y0 .
Recall Taylor expansion:
y(x + ∆x) = y(x) + y 0 (x)∆x +
y 00 (x)
(∆x)2 + · · · .
2
Let’s take the linear approximation at x, then
y(x + ∆x) ≈ y(x) + y 0 (x)∆x.
Euler’s method:
Given the initial condition y(x0 ) = y0 and the step size ∆x, for any k = 0, 1, 2, · · · , let yk = y(xk )
and
xk = xk−1 + ∆x = x0 + (k − 1)∆x.
Now we want to approximate yk : Replace x by xk in (1), then we get
yk+1 = yk + f (xk , yk )∆x.
(1)
3. (a) (5 points) Solve y 00 + 4y 0 + 13y = 0,
Answer.
y(0) = 2, y 0 (0) = 5.
The characteristic equation is r2 + 4r + 13 = (r + 2)2 + 9 = 0 with roots
r = −2 ± 3i.
(1pt)
y(t) = c1 e−2t cos 3t + c2 e−2t sin 3t.
(1pt)
The general solution is
Hence
y 0 (t) = (−2c1 + 3c2 )e−2t cos 3t + (−3c1 − 2c2 )e−2t sin 3t.
The initial conditions give
c1 + 0 = 2,
Thus c1 = 2, c2 = 3, and
(−2c1 + 3c2 ) + 0 = 5.
(2pt)
y(t) = 2e−2t cos 3t + 3e−2t sin 3t.
(1pt)
Remark 8. Let a, b and c be three constants, the characteristic equation of the differential equation
ay 00 + by 0 + cy = 0 is:
ar2 + br + c = 0.
Remark 9. Let a, b and c be three constants, the roots of ar2 + br + c = 0 are:
√
√
−b − b2 − 4ac
−b + b2 − 4ac
, and r2 =
.
r1 =
2a
2a
Then
1) If b2 − 4ac > 0, then r1 and r2 are two different real numbers, and
√
√
b2 − 4ac
−b
b2 − 4ac
−b
+
, and r2 =
−
.
r1 =
2a
2a
2a
2a
The general solution to ay 00 + by 0 + cy = 0 is:
y = C1 er1 x + C2 er2 x .
2) If b2 − 4ac = 0, then r1 = r2 = −
b
is a real number. The general solution to ay 00 + by 0 + cy = 0 is:
2a
−b
−b
y = C1 e 2a ·x + C2 xe 2a ·x .
3) If b2 − 4ac < 0, then r1 and r2 are two different complex numbers, and
√
√
b
4ac − b2
b
4ac − b2
r1 = − + i
, and r2 = − − i
2a
2a
2a
2a
The general solution to ay 00 + by 0 + cy = 0 is:
b
− 2a
·x
y = C1 e
cos
!
√
b
4ac − b2
· x + C2 e− 2a ·x sin
2a
!
√
4ac − b2
·x .
2a
(b) (5 points) Find a particular solution of
y 00 + 3y 0 = e−3t
using either the method of undetermined coefficients, or the method of variation of parameters.
Answer.
#1. (method of undetermined coefficients.) The characteristic equation is r2 + 3r = 0, with
roots −3, 0. Hence
yc (t) = c1 + c2 e−3t .
(2pt)
For f = e−3t , the initial guess is Y (t) = ae−3t . However ae−3t is part of yc and hence we set
Y (t) = ate−3t .
(1pt)
We have
Y 0 = (a − 3at)e−3t ,
Thus a = − 13 and
Y 00 = (−3a − 3a + 9at)e−3t ,
Y 00 + 3Y 0 = −3ae−3t .
(1pt)
1
Y (t) = − te−3t .
3
(1pt)
Remark 10. Undetermined Coefficients: Let a, b and c be constants, consider the equation:
ay 00 + by 0 + cy = f (x).
Let pn (x) and p̃n (x) be polynomials with degree n, a particular solution yp (x) to ay 00 + by 0 + cy = f (x)
can be taken as:
pn
f (x)
+ p̃n (x)emx sin(kx)
(x)emx cos(kx)
xα [q
n
yp (x)
mx
(x)e cos(kx) + q̃
n (x)e
mx sin(kx)]
where
• α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the solutions to the characteristic equation
ar2 + br + c = 0 of ay 00 + by 0 + cy = 0):
– If m + ki is not a root of ar2 + br + c = 0, then α = 0.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 has two different roots, then α = 1.
– If m + ki is a root of ar2 + br + c = 0 and ar2 + br + c = 0 only has one root m + ki, then α = 2.
• qn (x) and q̃n (x) are undetermined polynomials with degree n.
Answer #2. (method of variation of parameters.) The characteristic equation is r2 + 3r = 0, with roots
−3, 0. Hence two independent solutions are
y1 (t) = 1,
y2 (t) = e−3t .
(2pt)
A particular solution is of the form Y (t) = y1 (t)u1 (t) + y2 (t)u2 (t) with
u01 + e−3t u02 = 0,
Thus
0 u01 − 3e−3t u02 = e−3t .
1
u02 = − ,
3
1
u01 = e−3t .
3
(2pt)
We can solve u1 = − 19 e−3t and u2 = − 31 t, and hence
1
1
1
1
Y = 1(− e−3t ) + e−3t (− t) = − e−3t − te−3t .
9
3
9
3
(1pt)
Remark 11. Variation of Parameters: Let y1 (x) and y2 (x) be two linearly independent solutions of
y 00 + p(x)y 0 + q(x)y = 0, then a particular solution to y 00 + p(x)y 0 + q(x)y = f (x) can be
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x),
where u1 (x) and u2 (x) satisfy
y1 (x)u01 (x) + y2 (x)u02 (x) = 0,
y10 (x)u01 (x) + y20 (x)u02 (x) = f (x).
4. Consider a vibrating system described by the initial value problem
u00 + cu0 + 4u = cos 2t,
u(0) = 0,
u0 (0) = 2.
where c > 0 is the damping coefficient.
(a) (5 points) Find the steady periodic part of the solution (the part of the solution which remains as t → ∞)
of this problem, and find its amplitude. Do not find the transient part.
Answer.
The steady state is the particular solution of the form
U (t) = a cos 2t + b sin 2t.
Thus U 0 = 2b cos 2t − 2a sin 2t and U 00 = −4a cos 2t − 4b sin 2t. Substitution into the equation yields
c(2b cos 2t − 2a sin 2t) = cos 2t.
Thus a = 0, b =
1
2c ,
and
U (t) =
1
sin 2t.
2c
Its amplitude is
1
.
2c
Remark 12. When the damping constant c > 0, the general solution(also called transient solution) to
the homogeneous equation mx00 + cx0 + kx = 0 will go to 0 as t → ∞. The steady periodic solution of
mx00 + cx0 + kx = F0 cos(ωt) is the particular solution by using the method of undetermined coefficients.
Remark 13. The amplitude of the function A cos(ωt) + B sin(ωt) is
√
A2 + B 2 .
(b) (2 points) Let A(c) denote the maximum amplitude of the steady state solutions of the systems
u00 + cu0 + 4u = cos ωt,
u0 (0) = 2
u(0) = 0,
among all possible ω > 0. What happens to A(c) as c → 0+ ? Explain why.
Hint. You do not need to solve A(c) explicitly.
Answer.
A(c) goes to infinity, because A(c) ≥
1
2c
and limc→0+
1
2c
= ∞.
(c) (6 points) Find a particular solution of
y 00 + y =
Hint:
R
Answer.
tan t dt = − ln | cos t|,
R
1
,
cos t
−
π
π
<t< .
2
2
cot t dt = ln | sin t|
The homogeneous equation has two independent solutions
y1 (t) = sin t,
y2 (t) = cos t.
Using the method of variation of parameters, a particular solution is given by Y (t) = y1 (t)u1 (t) +
y2 (t)u2 (t), in which
y1 u01 + y2 u02 = 0,
y10 u01 + y20 u02 = f.
That is,
sin t u01 + cos t u02 = 0,
cos t u01 − sin t u02 =
1
.
cos t
Thus
u01 = 1,
u02 = − tan t
Z
u1 (t) = 1dt = t,
Z
u2 (t) = −
tan t dt = ln | cos t| = ln(cos t)
(−
π
π
<t< )
2
2
Hence a particular solution is
Y (t) = t sin t + cos t ln(cos t)
Remark 14. Variation of Parameters: Let y1 (x) and y2 (x) be two linearly independent solutions of
y 00 + p(x)y 0 + q(x)y = 0, then a particular solution to y 00 + p(x)y 0 + q(x)y = f (x) can be
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x),
where u1 (x) and u2 (x) satisfy
y1 (x)u01 (x) + y2 (x)u02 (x) = 0,
y10 (x)u01 (x) + y20 (x)u02 (x) = f (x).
5. (8 points) Use the Laplace transform to solve the system x00 (t) + 2x0 (t) + 2x(t) = 2 with initial conditions
x(0) = 0, x0 (0) = 0.
Answer.
Writing X(s) for the Laplace transform of x(t), we obtain
(s2 + 2s + 2)X(s) =
2
s
and therefore
X(s) =
2
.
s(s2 + 2s + 2)
By partial fraction decomposition, we have
X(s) =
s2
1
−s − 2
+ .
+ 2s + 2 s
By completion of the square, this can be put in a convenient form to later apply the first shifting formula:
X(s) =
−(s + 1)
1
1
−
+ .
(s + 1)2 + 1 (s + 1)2 + 1 s
Taking inverse Laplace transforms and applying the said shifting formula, we obtain
n s o
n 1 o
x(t) = −e−t L−1 2
(t) − e−t L−1 2
(t) + 1
s +1
s +1
= −e−t (cos t + sin t) + 1.
Remark 15. Let f (t) be a function on [0, ∞), then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
Z ∞
L[f (t)](s) =
f (t)e−st dt, for all s > 0.
0
II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as:
L−1 [F (s)](t) = f (t),
for all t > 0.
Remark 16. Let a be a real constant, and the Heaviside function is defined as:
1, if t ≥ a,
u(t − a) =
0, if x < a.
Notice that when a = 0, we know that u(t − 0) = 1 for all t ≥ 0; when a = ∞, we define u(t − ∞) := 0 for all
t ≥ 0.
Remark 17. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z t
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ.
0
Remark 18. Dirac’s delta function: For any continuous function f (t) on (−∞, ∞), we have
Z ∞
δ(t)f (t) dt = f (0).
−∞
Remark 19. The followings hold:
I. Transforms of derivatives:
L[f 0 ](s) = sL[f ](s) − f (0)
L[f 00 ](s) = s2 L[f ](s) − sf (0) − f 0 (0).
II. First Shifting Property:
L e−at f (t) (s) = L[f ](s + a)
L−1 [F (s + a)](t) = e−at L−1 [F (s)](t).
III. Second Shifting Property:
L[u(t − a)f (t − a)](s) = e−as L[f (t)](s)
L−1 e−as F (s) = u(t − a)L−1 [F (s)](t − a).
In practice, we can use the second shifting property in the following way:
L[u(t − a)f (t)](s) = e−as L[f (t + a)](s).
IV. Transform of Integrals:
Z
L
t
L[f (t)](s)
f (τ ) dτ (s) =
s
0
Z t
F (s)
L−1
(t) =
L−1 [F (s)](τ ) dτ
s
0
V. Transform of Convolution:
L[(f ∗ g)(t)](s) = L[f (t)](s) · L[g(t)](s)
L−1 [F (s) · G(s)] (t) = L−1 [F (s)] ∗ L−1 [G(s)](t).
VI. Transform of Dirac delta:
L[δ(t − a)](s) = e−as
L−1 [e−as ](t) = δ(t − a).
Remark 20. When dealing with the inverse Lapace tranform of a rational function, you should know the
completing square and partial fraction decomposition. For the completing square:
"
#
2
2
b
c
b
c
b
as2 + bs + c = a s2 + s +
=a s+
− 2+
.
a
a
2a
4a
a
6. (a) (6 points) Solve the system x00 (t) + 4x(t) = δ(t − 2) with initial conditions x(0) = 0 and x0 (0) = 0.
Answer.
Writing X(s) = L{x(t)} and taking Laplace transforms, we obtain
(s2 + 4)X(s) = e−2s
⇒
X(s) =
e−2s
.
s2 + 4
By the second shifting formula, we have therefore
1 o
(t − 2) · u(t − 2)
s2 + 4
= 21 sin(2(t − 2)) · u(t − 2).
x(t) = L−1
n
(b) (6 points) Find the Laplace transform of


1
f (t) = t


0
Answer.
if t < 1
if 1 ≤ t < 2
if t ≥ 2.
We can rewrite the piecewise definition of f using Heaviside functions:
f (t) = 1 · (1 − u(t − 1)) + t · u(t − 1) − u(t − 2) + 0 · u(t − 2)
= 1 + (t − 1)u(t − 1) − tu(t − 2)
= 1 + (t − 1)u(t − 1) − (t − 2)u(t − 2) − 2u(t − 2).
This last expression makes it easier to apply the second shifting formula:
2e−2s
1
+ e−s L{t}(s) − e−2s L{t}(s) −
s
s
1 e−s
2
−2s 1
= +
−e
+
.
s
s
s2
s
F (s) =
Remark 21. Let a0 = 0 < a1 < a2 < a3 < · · · < an


 f1 (t),


f2 (t),



 f3 (t),
f (t) =
..

.




 fn (t),


fn+1 (t),
< ∞, if
if 0 ≤ t < a1 ,
if a1 ≤ t < a2 ,
if a2 ≤ t < a3 ,
..
.
if an−1 ≤ t < an ,
if t ≥ an ,
then,
f (t) =
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 [u(t − an ) − u(t − ∞)]
k=1
= f1 (t)[1 − u(t − a1 )] +
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 (t)u(t − an ).
k=2
Notice that when a = 0, we know that u(t − 0) = 1 for all t ≥ 0; when a = ∞, we define u(t − ∞) := 0 for all
t ≥ 0.
7. (a) (7 points) Find general solution of
dx
= Ax,
dt
A=
p 4
−1 p
,
assuming that p is real.
Answer. We first find the eigenvalues of A by solving det(A − λI) = (p − λ)2 + 4 = 0. This gives two
complex conjugated eigenvalues λ1,2 = p ± 2i. The eigenvector that corresponds to λ1 = p + 2i is
−2i 4
0
2
v1 =
⇒ v1 =
.
−1 −2i
0
i
The general solution is then given by
2 2 (p+2i)t
e(p+2i)t ,
e
+ C2 Im
x = C1 Re
i
i
which can be simplified to
pt
x = C1 e
2 cos(2t)
− sin(2t)
pt
+ C2 e
2 sin(2t)
cos(2t)
.
Remark 22. To find the general solution to ~x0 = A~x, we have the following three cases:
• Let A be a 2 × 2 matrix with two distinct real eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then the general solution to ~x0 = A~x is:
~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 .
• Let A be a 2×2 matrix with two distinct complex eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then λ2 = λ1 , ~v2 = ~v1 , and the general solution to ~x0 = A~x
is:
~x = C1 Re eλ1 t~v1 + C2 Im eλ1 t~v1 .
• Let A be a 2 × 2 matrix with the same real eigenvalue λ and ~v be the eigenvector corresponding to
λ, then ~x(t) = eλt~v is a solution to ~v 0 = A~x.
0
1
are two linearly independent eigenvectors corresponding to
and
– If λI2 − A = 0, then
1
0
λ, that is, λ is a complete eigenvalue of A. Hence the general solution to ~x0 = A~x is
1
0
~x(t) = eλt
+ eλt
.
0
1
– If λI2 − A 6= 0, that is, λ is a defective eigenvalue of A, then another linearly independent
solution ~y to ~x0 = A~x has the form ~y (t) = eλt (t~v + w)
~ for some vector w
~ to be determined. That
is, ~y (t) = t~x(t) + eλt w.
~ So we get
~y 0 (t) = ~x(t) + t~x0 (t) + λeλt w
~
= ~x(t) + tA~x(t) + λeλt w
~
= A~y (t)
= tA~x(t) + eλt Aw.
~
Then ~x(t) + λeλt w
~ = eλt Aw,
~ that is, eλt~v + λeλt w
~ = eλt Aw,
~ so w
~ is a solution to ~v + λw
~ = Aw,
~
that is,
(A − λI2 )w
~ = ~v .
In this case, we call w
~ a generalized eigenvector of A.
(b) (3 points) Describe the behaviour of the system (do not draw phase portrait) for all possible real values
of p.
Answer. Since the eigenvalues are λ1,2 = p ± 2i and Re(λ1,2 ) = p is real, there are only three cases. If p > 0,
then the system behaves like a spiral source. If p < 0, then the system behaves like a spiral sink. If p = 0,
then the system behaves like a center.
a b
be a 2 × 2 matrix, λ1 and λ2 be two nonzero different eigenvalues of A.
c d
Notice that det(A − λI2 ) = λ2 − (a + d)λ + (ad − bc). Since λ1 6= 0 and λ2 6= 0, then det A 6= 0, which implies
that the origin (0, 0) is the only critical point to the system ~x0 = A~x.
Remark 23. Let A =
For the behaviors of solutions to ~x0 = A~x near the origin (0, 0), we have the following six cases:
Eigenvalues
λ1 , λ2
λ1 , λ2
λ1 , λ2
λ1 , λ2
λ1 , λ2
λ1 , λ2
are
are
are
are
are
are
real and
real and
real and
complex
complex
complex
both positive
both negative
opposite signs
with positive real part
with negative real part
with zero real part
Type of (0,0)
Stability of (0,0)
source
sink
saddle point
spiral source
spiral sink
center point
unstable
asymptotically stable
unstable
unstable
asymptotically stable
stable
8. (12 points) Solve
dx
= Ax −
dt
0
3t + 2
,
x(0) =
0
1
,
A=
0 1
3 2
.
Answer.
We start by finding the complimentary solution. The eigenvalues of matrix A can be found from
−λ(2 − λ) − 3 = 0, which gives λ1 = 3 and λ2 = −1. The corresponding eigenvectors are
−3 1
0
1
v1 =
⇒ v1 =
,
3 −1
0
3
1 1
0
1
v2 =
⇒ v2 =
.
3 3
0
−1
The complimentary solution is then
xc = C1
1
3
3t
e + C2
1
−1
e−t .
Method 1. By undetermined coefficients, we seek for the particular solution in the form xp = at + b, where
a and b are constant vectors with unknown components. Substituting xp into the equation gives
0
0
a = Aat + Ab −
−
t,
2
3
which implies that
a = Ab −
0
2
,
Aa =
0
3
.
Computing the the inverse of A gives
A
1
=
3
−1
−2 1
3 0
,
which can be used to find
a = A−1
0
3
=
1
0
b = A−1
,
1
2
=
0
1
.
So, the particular solution is
xp =
t
1
.
Remark 24. Undetermined coefficients: Let A be a 2 × 2 matrix,, consider the system:
~x0 = A~x + f~(t).
Let p~n (t) and ~qn (t) be “vector coefficients” polynomials with degree n. If f~(t) has the form:
p~n (t)emt cos(kt) + ~qn (t)emt sin(kt),
then a particular solution ~xp (t) to ~x0 = A~x + f~(t) can be take as:
~xp (t) = p~n+α (t)emt cos(kt) + ~qn+α (t)emt sin(kt),
where
• α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the eigenvalues to A):
– If m + ki is not an eigenvalue of A, then α = 0.
– If m + ki is an eigenvalue of Aand A has two different eigenvalues, then α = 1.
– If m + ki is an eigenvalue of A and A has only one eigenvalue, then α = 2.
• p~n+α (t) and ~qn+α (t) are undetermined “vector coefficients” polynomials with degree n + α.
Method 2. To use variation of parameters, we first need to find fundamental matrix solution. Using the
obtained complimentary solution, one has
3t
1 e−3t e−3t
e
e−t
−1
X=
,
X =
.
3e3t −e−t
3et −et
4
Particular solution can be found using
3t
R
Z
1
e
e−t
0
(−3t
− 2)e−3t dt
−1
R
dt =
.
xp = X X
−3t − 2
(3t + 2)et dt
4 3e3t −e−t
The integrals can be calculated as
R
(−3t
− 2)e−3t dt
R
(3t + 2)et dt
=
(t + 1)e−3t
(3t − 1)et
.
Finally, the particular solution is
1
xp =
4
e3t
e−t
3t
3e
−e−t
(t + 1)e−3t
(3t − 1)et
=
t
1
.
The general solution is
x = C1
1
3
1
3
3t
e + C2
1
−1
e
−t
0
1
+
t
1
0
1
.
Using the initial condition gives
x(0) = C1
+ C2
1
−1
+
=
,
which implies that C1 = C2 = 0. Finally, the solution is
t
.
x=
1
y1 (t)
z1 (t)
and ~z(t) =
Remark 25. Let A be a 2 × 2 matrix, and ~y (t) =
be two linearly independent
y2 (t)
z2 (t)
y1 (t) z1 (t)
, then X(t) is a fundamental matrix, and
solutions to ~x0 = A(t)~x, let X(t) = [~y (t) ~z(t)] =
y2 (t) z2 (t)
X 0 (t) = A(t)X(t). Hence the general solution to ~x0 = A(t)~x is:
C1
y1 (t) z1 (t)
C1
~
~x(t) = C1 ~y (t) + C2~z(t) = [~y (t) ~z(t)]
=
= X(t)C.
C2
y2 (t) z2 (t)
C2
Remark 26. Let X(t) be any fundamental matrix of ~x0 = A(t)~x. A particular solution to ~x0 = A~x + f~ is
Z t
~xp (t) = X(t) [X(s)]−1 f~(s) ds.
The solution to the initial value problem ~x0 = A~x + f~ and ~x(0) = ~x0 is
Z t
−1
~x(t) = X(t)[X(0)] ~x0 + X(t) [X(s)]−1 f~(s) ds.
0
9. Consider the nonlinear system
dx
y2
= x2 −
+ 1,
dt
2
dy
= −4x − 2y.
dt
(2)
(a) (5 points) Find the equilibria (critical points) for the system (2).
2
Answer.
The equilibria are solutions of the equations x2 − y2 + 1 = 0 and −4x − 2y = 0. From the
second equation, we obtain y = −2x. Substituting this into the first equation, we have −x2 + 1 = 0.
Hence the equilibria are (1, −2) and (−1, 2).
Remark 27. Consider the autonomous system ~x0 = f~(~x), let ~a be a constant vector, we say that ~a is a
critical point of the autonomous system ~x0 = f~(~x) if f~(~a) = 0. In this case, we say that ~x(t) ≡ ~a is an
equilibrium solution to the system ~x0 = f~(~x).
(b) (6 points) Find the Jacobian (partial derivative) matrix for the system (2), compute the linearized system
at each equilibrium, and compute the eigenvalues for each of the coefficient matrices.
Answer.
The Jacobian matrix is
2x −y
−4 −2
.
u0
2
2
u
linearized system at (1, −2) is
=
.
0
v
−4 −2
v
characteristic equation is λ2 + 4 = 0 and the eigenvalues are ±2i.
0 u
−2 −2
u
linearized system at (−1, 2) is
=
.
0
v
−4 −2
v
characteristic equation is λ2 + 4λ − 4 = 0, and the eigenvalues are
√
√
−4 ± 32
= −2 ± 2 2.
2
The
The
The
The
Remark 28. The Jacobian matrix of a vector function
fx (a, b) fy (a, b)
gx (a, b) gy (a, b)
f (x, y)
g(x, y)
at the point (a, b) is the matrix:
x0 = f (x, y)
, that is, f (x0 , y0 ) =
y 0 = g(x, y)
g(x0 , y0 ) = 0. Let u = x − x0 and v = y − y0 , the linearization at (x0 , y0 ) of the autonomous system
0
x = f (x, y)
is:
y 0 = g(x, y)
0 u
fx (x0 , y0 ) fy (x0 , y0 )
u
.
=
v
gx (x0 , y0 ) gy (x0 , y0 )
v0
Remark 29. Let (x0 , y0 ) be a critical point to the autonomous system
(c) (4 points) Identify the equilibrium of (2) at which the linearized system (not system (2)) has a center.
Classify the other equilibrium and indicate whether it is (asymptotically) stable or unstable in system
(2).
Answer.
The eigenvalues of the coefficient matrix for the linearized system at equilibrium (1, −2) are
are purely imaginary. Therefore the linearized system at (1, −2) has a center.
The coefficient matrix for the linearized system at equilibrium (−1, 2) has a positive eigenvalue and a
negative eigenvalue. Hence, the equilibrium is a saddle, which is unstable, in system (2).
Remark 30. Let (x0 , y0 ) be a critical point of the system
x0 = f (x, y)
, that is, f (x0 , y0 ) = g(x0 , y0 ) =
y 0 = g(x, y)
0, then
I. We say that the critical point (x0 , y0 ) is isolated if it is the only critical point in a small “neighborhood” of (x0 , y0 ).
0
x = f (x, y)
II. We say that the system
is almost linear at critical point (x0 , y0 ) if the critical
y 0 = g(x, y)
f (x, y)
point ~a is isolated, and the Jacobian matrix of
at (x0 , y0 ) is invertible.
g(x, y)
x0 = f (x, y)
, that is, f (x0 , y0 ) =
Remark 31. we assume that (x0 , y0 ) is a critical point of the system
y 0 = g(x, y)
f (x, y)
at (x0 , y0 ) is invertible. Also we assume that the
g(x0 , y0 ) = 0, and the Jacobian matrix of
g(x, y)
f (x, y)
at (x0 , y0 ) has two differential eigenvalues λ1 and λ2 . The following table
Jacobian matrix of
g(x, y)
0
x = f (x, y)
near this isolated critical point (x0 , y0 ):
shows the behavior of the system
y 0 = g(x, y)
Eigenvalues of Jacobian at (x0 , y0 )
Type of (x0 , y0 )
Stability of (x0 , y0 )
λ1
λ1
λ1
λ1
λ1
source
sink
saddle
spiral source
spiral sink
unstable
asymptotically stable
unstable
unstable
asymptotically stable
and
and
and
and
and
λ2
λ2
λ2
λ2
λ2
are
are
are
are
are
real and
real and
real and
complex
complex
both positive
both negative
opposite signs
with positive real part
with negative real part