Math 2280
Summer 2006
Final
You will have two hours to complete this test. You may not use a calculator, computer, or any other electronic
device on this test. Please be sure to show as much of your work as possible, as I will determine your score
on a given problem from the work and not necessarily by the answer. Please be neat so that I can grade your
answers easily. Also, if there is a possibility of a complex valued solution to an ODE, give a real valued solution
only.
1. Find an implicit solution to the ODE
√
1+ x
dy
=
√ .
dx
1+ y
We simply separate the variables and integrate:
Z
Z
√
2 3
√
2 3
(1 + y) dy = (1 + x) dx ⇒ y + y 2 = x + x 2 + C.
3
3
2. Find a solution to
y ′ + 2xy = x
that satisfies y(0) = −2.
This is a first order linear equation, so we introduce the integrating factor
Z
2
ρ(x) = exp 2x dx = ex .
Multiplying through by ρ(x), we find that
x2 ′
x2
e y + e 2xy = xe
x2
Z
h 2 i
2
x
x2
x2
⇒ Dx e y = xe
⇒ e y = xex dx.
To evaluate the integral, we let u = x2 , then du = 2x dx. This gives that
ex
2
y
=
2
1
1 x2
e + C ⇒ y = + Ce−x .
2
2
The initial condition implies that
1
5
+ C ⇒ − = C.
2
2
−2 =
Thus, the solution is
y(x) =
1 5 −x2
− e
.
2 2
3. Find a general solution to the equation
2y ′′ + 3y ′ = 0.
(Primes denote differentiation with respect to x.)
The characteristic equation for the ODE is
3
2r2 + 3r = 0 ⇒ r(2r + 3) = 0 r = 0 or r = − .
2
This gives the solution
3
y(x) = c1 + c2 e− 2 x .
1
4. Find a general solution to the equation
9y (3) + 12y ′′ + 4y ′ = 0
(Primes denote differentiation with respect to x.)
The characteristic equation for the ODE is
2 2
9r3 + 12r2 + 4r = 0 ⇒ r(9r2 + 12r + 4) = 0 ⇒ r = 0 or (3r + 2)2 = 0 ⇒ r = 0, − , − .
3 3
This gives us the solution
2
2
y(x) = c1 + c2 e− 3 x + c3 xe− 3 x .
5. Solve the initial value problem
y ′′ − 6y ′ + 25y = 0, y(0) = 3, y ′ (0) = 1
(Primes denote differentiation with respect to x.)
The characteristic equation for the ODE is
2
r − 6r + 25 = 0 ⇒ r =
6±
p
√
36 − 4(1)(25)
6 ± −64
=
= 3 ± 4i.
2
2
This gives the general solution
y(x) = c1 e3x cos 4x + c2 e3x sin 4x ⇒ y ′ (x) = c1 (−4e3x sin 4x + 3e3x cos 4x) + c2 (4e3x cos 4x + 3e3x sin 4x).
Evaluating these equations at the initial point tells us that
3 = c1
1 = 3c1 + 4c2
⇒
c1 = 3
.
c2 = −2
This gives the final solution
y(x) = 3e3x cos 4x − 2e3x sin 4x.
6. Find the general solution of the ODE
y (4) + 5y ′′ − 36y = 0.
(Primes denote differentiation with respect to x.)
The characteristic equation for this ODE is
r4 + 5r2 − 36 = 0
If we let u = r2 , it becomes
u2 + 5u − 36 = 0 ⇒ (u − 4)(u + 9) = 0.
Thus, r = ±2 or r = ±3i. This gives the general solution
y(x) = c1 e2 x + c2 e−2x + c3 cos(3x) + c4 sin(3x).
7. Find the general solution to the system
x′1 = x1 − 5x2
.
x′2 = x1 − x2
The eigenvalues of the coefficient matrix satisfy
1−λ
−5 0=
= (1 − λ)(−1 − λ) + 5 = λ2 + 4 ⇒ λ = ±2i.
1
−1 − λ 2
An eigenvector corresponding to the eigenvalue λ = 2i is found by the calculation
1 − 2i
−5
1 −1 − 2i
1 + 2i
1
∼
⇒ v1 = (1 + 2i)v2 → ~v1 =
⇒ ~a =
,
1
−1 − 2i
0
0
1
1
~b =
2
0
.
This gives us the general solution
~x(t) = (c1 cos 2t + c2 sin 2t)
1
1
1
4
+ (c2 cos 2t − c1 sin 2t)
2
0
.
8. Find the general solution to the system
~x′ =
−4
9
~x.
The eigenvalues of the coefficient matrix must satisfy
1−λ
−4 0=
= (1 − λ)(9 − λ) + 16 = λ2 − 10λ + 25 = (λ − 5)2 ,
4
9−λ so we have a repeated eigenvalue. By the following calculation, we find the eigenvector(s) corresponding to
λ=5
−4 −4
1 1
−1
∼
⇒ v1 = −v2 ⇒ ~v1 =
.
4
4
0 0
1
There is only a one dimensional

 −4 −4
4
4
space of eigenvectors, so we must find a generalized eigenvector:

 
..
1 .. 1
. −1   1 1 . 4 
4
.
⇒
v
=
−v
⇒
~
v
=
∼
1
2
2
.
..
0
0 0 .. 0
. 1
This gives the general solution
~x(t) = c1 e
5t
−1
1
+ c2 e
5t
1
4
0
+t
−1
1
.
9. Find the matrix exponential for the system
x′1 = 3x1
.
x′2 = 4x2
The easiest way to do this is to note that
1 9t2
3t 0
eAt = I +
+
0
0 4t
2!
0
16t2
+ ··· =
e3t
0
0
e4t
1 −2
−2 −3
.
10. Find the linearization of the system
dx
= x − 2y + 3xy
dt
dy
= −2x − 3y − x2 − y 2
dt
about the origin and classify it as a node, center, etc.
We have that
J(x, y) =
The eigenvalues must satisfy
1 + 3y −2 + 3x
−2 − 2x −3 − 2y
1−λ
−2
0 = −2 −3 − λ
⇒ J(0, 0) =
.
= (1 − λ)(−3 − λ) − 4 = λ2 + 2λ − 7.
3
The quadratic formula gives that
λ=
−2 ±
p
√
√
4 − 4(1)(−7)
−2 ± 32
=
= −1 ± 2 2.
2
2
Thus, the origin is a saddle.
11. Using variation of parameters and the matrix exponential as the fundamental matrix, find the solution of the
initial value problem
~x′ = A~x + ~b, ~x(0) = ~x0 ,
where ~b is a constant vector. Be sure to simplify your answer.
If our fundamental matrix is eAt , then any solution to the homogeneous problem can be written as
~x(t) = eAt~c
for some choice of vector ~c. Therefore, we let
~xp (t) = eAt ~u(t) ⇒ ~x′p (t) = eAt ~u′ (t) + AeAt ~u(t).
Substituting ~xp (t) into the original equation gives that
eAt ~u′ (t) + AeAt ~u(t) = AeAt ~u(t) + ~b ⇒ ~u′ (t) = e−At~b.
This gives that
~u(t) =
so
Z
e−At~b dt = −A−1 e−At~b,
~xp (t) = eAt (−A−1 e−At~b) = −A−1~b.
The general solution is
~x(t) = eAt~c − A−1~b.
Applying the initial condition gives us that
~x0 = ~c − A−1~b ⇒ ~c = ~x0 + A−1~b.
The solution to the problem is
~x(t) = eAt (~x0 + A−1~b) − A−1~b = eAt ~x0 + (eAt − I)A−1~b.
12. Using the definition, find the Laplace transform of f (t) = t2 and give its domain.
We have that
F (s) =
Z
∞
e−st t2 dt.
0
To evaluate this integral, we let u = t2 and dv = e−st dt and integrate by parts to obtain
2
∞
Z
t −st
2 ∞ −st
F (s) = − e
+
te
dt.
s
s 0
0
The first term is zero if we assume that s > 0. We must integrate by parts again, this time with u = t and
dv = e−st dt. This gives that
∞
Z
2
1 −st
2 ∞ −st
2 −st ∞
2 1
2
= 3 , s > 0.
+ 2
te
F (s) =
= 2
e
dt = 2 − e
0
s
s 0
s
s
s s
s
0
13. Using Laplace transforms, solve the initial value problem
x′′ + 4x = cos t,
x(0) = 0 = x′ (0).
4
The transformed equation becomes
s2 X(s) + 4X(s) =
s2
s
s
⇒ X(s) = 2
.
+1
(s + 1)(s2 + 4)
We use partial fractions to decompose the right hand side of the last equation above:
As + B
Cs + D
s
= 2
+ 2
(s2 + 1)(s2 + 4)
s +1
s +4
⇒ s = (A + C)s3 + (B + D)s2 + (4A + C)s + 4B + D ⇒ A + C = 0,
These equations are easily solved to give A =
X(s) =
1
3,
B = 0, C =
− 31 ,
B + D = 0,
and D = 0. Thus
s
1
1
s
−
⇒ x(t) = cos t − cos 2t.
2
+ 1) 3(s + 4)
3
3
3(s2
14. Use the convolution property to find the inverse Laplace transform of
F (s) =
We have that
F (s) =
Thus,
1
.
s(s − 3)
1
1
1
= ·
= L{1} · L{e3t }.
s(s − 3)
s s−3
L−1 {F (s)} = (1 ∗ e3t ) =
Z
t
e3τ dτ =
0
1 3τ
e
3
t
0
=
1 3t 1
e − .
3
3
15. Find the inverse Laplace transform of
F (s) = ln s.
F (s) = ln s ⇒ F ′ (s) =
1
1
= L{1} ⇒ − = L−1 {ln s}.
s
t
FYI
L{sin kt} =
L{cos kt} =
L{1} =
5
k
s2 +k2
s
s2 +k2
1
s
4A + C = 1,
4B + D = 0