Math 2280
Summer 2009
Final Exam
You will have two hours to complete this test. You may not use a calculator 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 from the answer. Please be neat so that I can grade your answers easily.
Some Useful Laplace Transforms
f(t)
t (n ≥ 0)
eat
cos kt
sin kt
F(s)
n!
sn+1
1
s−a
s
s2 +k2
k
s2 +k2
n
1. Find the solution to the initial value problem
x
If we rewrite this as
x
dy
− y = 2x2 y,
dx
dy
= 2x2 y + y = y(2x2 + 1),
dx
then
dy
=
y
Z
On the left we have
and on the other side we have
2x2 + 1
dx =
x
Z
2x2 + 1
dx.
x
dy
= ln y,
y
Z
Z
y(1) = 1.
Z 1
2x +
dx = x2 + ln x + C.
x
Solving for y, we find that
y(x) = ex
2
+ln x+C
2
= Cxex .
Upon applying the initial condition, we find
1 = y(1) = Ce ⇒ C = e−1 ⇒ y(x) = xex
2
−1
.
2. Find the general solution to the problem
y ′ + 2xy = x.
This equation is first-order linear, so we find an integrating factor:
Z
2
ρ(x) = exp
2x dx = ex .
Multiplying through the equation by ρ(x), we obtain
2
2
2
ex y ′ + 2xex y = xex .
2
2
2
We now recognize that Dx [ex y] = ex y ′ + 2xex y, so that taking an antiderivative of both sides yields
Z
2
2
ex y = xex dx.
1
If we let u = x2 so that du = 2x dx, then the integral becomes
Z
1
1 u
e du = eu + C.
2
2
Therefore,
2
ex y =
2
1 x2
1
e + C ⇒ y = + Ce−x .
2
2
3. Use Euler’s method with stepsize h = 1/2 to find approximations to y(1/2) and y(1) where y is the solution
to the IVP
y ′ + 3y 2 = x, y(0) = 1.
The iteration formula for Euler’s method is
yn+1 = yn + hf (xn , yn ),
and for this problem
f (x, y) = x − 3y 2 , x0 = 0, y0 = 1,
Therefore,
1
1
y(1/2) ≈ y1 = y0 + hf (x0 , y0 ) = 1 + [−3] = − ,
2
2
"
2 #
1 1 1
1 1
1
5
y(1) ≈ y2 = y1 + hf (x1 , y1 ) = − +
=− − =− .
−3 −
2 2 2
2
2 8
8
4. Find the general solution to the ODE
y ′′ + y = sin x.
The characteristic equation is
r2 + 1 = 0,
so r = ±i. This means that
yc (x) = c1 cos x + c2 sin x.
It is now clear that f (x) = sin x is part of the complimentary solution, so we take as our guess for a particular
solution
yp (x) = Ax sin x + Bx cos x.
Then
yp′ (x) = A sin x + Ax cos x + B cos x − Bx sin x,
yp′′ (x) = A cos x + A cos x − Ax sin x − B sin x − B sin x − Bx cos x = 2A cos x − Ax sin x − 2B sin x − Bx cos x.
Substituting into the equation, we must have
sin x = 2A cos x − Ax sin x − 2B sin x − Bx cos x + Ax sin x + Bx cos x = 2A cos x − 2B sin x,
so 2A = 0 and −2B = 1. Therefore
The general solution is
1
yp (x) = − x cos x.
2
1
y(x) = c1 cos x + c2 sin x − x cos x.
2
5. Find the solution to the initial value problem
y ′′ − y ′ = 0,
y(0) = 2, y ′ (0) = 1.
2
The characteristic equation is
r2 − r = 0,
so r = 0 or r = 1, and therefore y(x) = c1 ex + c2 , so y ′ (x) = c1 ex . The initial conditions imply that
and 1 = y ′ (0) = c1 .
2 = y(0) = c1 + c2
Therefore, c1 = c2 = 1, and the solution is
y(x) = ex + 1.
6. Find the general solution to the ODE
y ′′ − 4y ′ + 4y = 0.
The characteristic equation is
r2 − 4r + 4 = 0 ⇒ (r − 2)2 = 0 ⇒ r = 2, 2.
This implies that
y(x) = c1 e2x + c2 xe2x .
7. Find a particular solution to the problem
~x′ =
1
−1
1
3
~x +
3e2t
0
R
using the variation of parameters formula ~xp (t) = Φ(t) Φ−1 (t)f~(t) dt.
As always, first we must find the complimentary solution. The characteristic equation is
1−λ
1 = (1 − λ)(3 − λ) + 1 = 3 − λ − 3λ + λ2 + 1 = λ2 − 4λ + 4 = (λ − 2)2 ,
0=
−1
3−λ so λ = 2, 2. We need to find out how many linearly independent eigenvectors correspond to λ = 2. We have
−1 1
1 −1
1
A − 2I =
∼
⇒ a = b ⇒ ~v1 =
.
−1 1
0 0
1
We have only one linearly independent eigenvector, so we must find a generalized eigenvector. We do this by
solving the system (A − λI)~v2 = ~v1 , which has the augmented matrix

 

..
..
 −1 1 . 1  ∼  1 −1 . −1  ⇒ a = b − 1 ⇒ ~v2 = −1 .
.
.
0
−1 1 .. 1
0 0 .. 0
Therefore, the complimentary solution is
~xc (t) = c1 e
2t
1
1
+ c2 e
e2t
e2t
2t
−1
0
+t
1
1
.
A fundamental matrix for the system is
Φ(t) =
te2t − e2t
te2t
.
We have that det(Φ(t)) = te4t − (te4t − e4t ) = e4t , so
1
te2t e2t − te2t
te−2t
=
Φ−1 (t) = 4t
2t
2t
−e
e
−e−2t
e
3
e−2t − te−2t
e−2t
and
Φ
−1
(t)f~(t) =
Upon integrating, we find
Z
Φ
−1
te−2t
−e−2t
e−2t − te−2t
e−2t
(t) dt =
Z Finally, we multiply by Φ(t) to get
2t
3 2 e
te2t − e2t
2t
~xp (t) =
=
e2t
te2t
−3t
3t
−3
dt =
3e2t
0
3 2
2t
−3t
3 2 2t
2 2t
2t
2 t e − 3t e + 3te
3 2 2t
2 2t
2 t e − 3t e
=
3t
−3
=
.
.
− 23 t2 e2t + 3te2t
− 23 t2 e2t
.
8. Find the general solution to the system
′
~x =
1 8
0 3
~x.
This matrix is upper triangular, so its eigenvalues are clearly 1 and 3. In order to find the solution, we need to
write down the eigenvectors corresponding to these eigenvalues. For λ = 1, we have
0 8
0 1
1
A−I =
∼
⇒ b = 0 ⇒ ~v1 =
.
0 2
0 0
0
For λ = 3, we have
A − 3I =
Therefore, the solution is
−2 8
0 0
∼
~x(t) = c1 e
1
0
t
−4
0
1
0
0
1
⇒ a = 4b ⇒ ~v2 =
+ c2 e
3t
4
1
4
1
.
.
9. Find the general solution to the problem
′
~x =
−1
0
~x.
The eigenvalues of the coefficient matrix must satisfy
−λ −1 = λ2 + 1,
0=
1 −λ so λ = ±i. Since we have complex eigenvalues, we choose one of them (say λ = i) and find a corresponding
eigenvector:
−i −1
1 −i
i
0
1
A − iI =
∼
⇒ a = ib ⇒ ~v1 =
=
+
i = ~a + ~bi.
1 −i
0 0
1
1
0
Recall that for a complex eigenvalue/eigenvector, the corresponding part of the solution can be written as
eλt~v = e(p+qi)t (~a + ~bi) = ept (cos qt + i sin qt)(~a + ~bi) = ept (cos qt~a − sin qt~b) + iept (cos qt~b + sin qt~a).
Here we use the real and imaginary parts of eit~v1 as our linearly independent solutions, so we have
1
0
0
−c1 cos t + c2 sin t
1
+ sin t
=
− sin t
.
+ c2 cos t
~x(t) = c1 cos t
0
1
1
c1 cos t + c2 sin t
0
10. Use the definition of the matrix exponential and the fact that
2 −2
A=
2 −2
4
is nilpotent to find the matrix exponential for the system ~x′ = A~x.
The definition of the matrix exponential is
eAt =
∞
X
Ak tk
k=0
In our case we have that
2
A =
so
e
At
At
=I+
=
1!
2
2
1
0
0
1
k!
−2
−2
2
2
−2
−2
2t
2t
−2t
−2t
+
.
=
=
0
0
0
0
,
2t + 1
2t
−2t
−2t + 1
.
11. For the system
dx
dt
dy
dt
= 14x − 21 x2 − xy
,
= 16y − 21 y 2 − xy
find all the fixed points, linearize the system about each fixed point, and tell the type (node, saddle, center, spiral,
etc.) of fixed point the linearization predicts. Then say whether the linearized fixed point is stable, unstable, or
asymptotically stable.
In order to have a fixed point, we must have
dx
1
1
= 0 ⇒ 14x − x2 − xy = 0 ⇒ x = 0 or 14 − x − y = 0,
dt
2
2
and we also must have
1
1
dy
= 0 ⇒ 16 − y 2 − xy = 0 ⇒ y = 0 or 16 − y − x = 0.
dt
2
2
Therefore, there are fixed points at (0, 0), (0, 32), (28, 0), and at the intersection of the two lines. We can find
the intersection via augmented matrix as follows:

 "

# 
..
..
.
1
..
28 ∼  1 0 . 12  ,
 2 1 . 14  ∼ 1 2
.
..
3
1
0 − 2 −12
1 2 . 16
0 1 .. 8
so the other fixed point is at (12, 8). The Jacobian for the system is
14 − x − y
−x
J(x, y) =
.
−y
16 − y − x
At (0, 0), we have
J(0, 0) =
14 0
0 16
⇒ λ1 = 14, λ2 = 16.
Thus, (0, 0) is an unstable node. At (0, 32), we have
−18
0
J(0, 32) =
⇒ λ1 = −18, λ2 = −16.
−32 −16
Thus (0, 32) is an asymptotically stable node. At (28, 0) we have
−14 −28
J(28, 0) =
⇒ λ1 = −14, λ2 = −12.
0
−12
Here we have another asymptotically stable node. Finally, at (12, 8), we have
−6 −12
J(12, 8) =
.
−8 −4
5
The eigenvalues satisfy
−6 − λ
−12
0 = −8
−4 − λ
= (−6 − λ)(−4 − λ) − 96 = 24 + 6λ + 4λ + λ2 − 96
= λ2 + 10λ − 72 ⇒ λ2 + 10λ + 25 = 72 + 25 ⇒ (λ + 5)2 = 97 ⇒ λ = −5 ±
√
97.
√
Since 97 > 5, we have one positive and one negative eigenvalue. Therefore, (12, 8) is a saddle point, which is
unstable.
12. Find the Laplace transform of the function defined by
3t
e , 0≤t≤1
f (t) =
.
0,
otherwise
By definition, we have that
Z ∞
Z
−st
F (s) =
e f (t) dt =
0
1
e
−st 3t
e
dt =
0
Z
1
e
(3−s)t
0
so
L{f (t)} =
1 (3−s)t
e
dt =
3−s
1
0
e3−s
1
,
=
−
3−s 3−s
e3−s − 1
.
3−s
13. Use Laplace transforms to solve the initial value problem
x′′ − 7x′ + 10x = 0, x(0) = 1, x′ (0) = −1.
Taking the Laplace transform of both sides, we have that
L{x′′ } − 7L{x′ } + 10L{x} = 0,
by the linearity of the Laplace transform. Applying the theorem about transforms of derivatives, this becomes
s2 L{x} − sx(0) − x′ (0) − 7sL{x} + 7x(0) + 10L{x} = 0
⇒ (s2 − 7s + 10)L{x} = s − 1 − 7 = s − 8
⇒ L{x} =
s2
s−8
s−8
=
.
+ 7s + 10
(s − 2)(s − 5)
In order to find the inverse Laplace transform of the right hand side, we first need to break it up using partial
fractions. We write
s−8
A
B
=
+
⇒ s − 8 = A(s − 5) + B(s − 2) = (A + B)s − 5A − 2B
(s − 2)(s − 5)
s−2 s−5
⇒ A + B = 1, −5A − 2B = −8.
We can set this system up as an augmented matrix:
 

..
.
1
1
1
∼ 1 1

..
0 3
−5 −2 . −8
so A = 2 and B = 1. We now have that
L{x} =
and upon consulting the table below, we find that
 
..
. 1   1
∼
..
0
. −3
2
−1
+
,
x−2 x−5
x(t) = 2e2t − e5t .
6

.
0 .. 2 
,
.
1 .. −1
14. Use Laplace transforms to solve the initial value problem
x′′ − 7x′ + 10x = e2t , x(0) = 0, x′ (0) = 1.
The left-hand side of the equation is the same as in the previous problem, so when we Laplace transform, we
get
s2 L{x} − sx(0) − x′ (0) − 7sL{x} + 7x(0) + 10L{x} = L{e2t }.
Upon rearranging, we find that
(s2 − 7s + 10)L{x} = 1 +
1
s−1
s−1
=
⇒ L{x} =
.
s−2
s−2
(s − 2)2 (s − 5)
Once again, we must use partial fractions before we can perform the inverse Laplace transform via our table. We
write
C
A
B
s−1
+
=
+
⇒ s−1
(s − 2)2 (s − 5)
s − 2 (s − 2)2
s−5
= A(s − 2)(s − 5) + B(s − 5) + C(s − 2)2 = A(s2 − 7s + 10) + B(s − 5) + C(s2 − 4s + 4)
= (A + C)s2 + (−7A + B − 4C)s + (6A − 5B + 4C).
Therefore, we have the equations
A + C = 0, −7A + B − 4C = 1, 10A − 5B + 4C = −1.
As an augmented matrix, this system has the form

 
..
.
1
0
1
.
0
1 .. 0

  1 0

 
.
.
 −7 1 −4 .. 1  ∼  0 1
3 .. 1

 
.
.
10 −5 4 .. −1
0 −5 −6 .. −1


  1 0
 
∼ 0 1
 
0 0
so A = −4/9, B = −1/3, and C = 4/9. We can now write
L{x} = −
 
.
1 .. 0   1
 
.
∼
3 .. 1 
  0
.
0
9 .. 4
0 0
1 0
0 1
..
. − 94
..
. − 13
..
4
.
9



,

1
4 1
4 1
1
+
−
.
9 s − 2 3 (s − 2)2
9s−5
Consulting the table and using the translation theorem, we see that
4
1
4
x(t) = − e2t − te2t + e5t .
9
3
9
15. Find the inverse Laplace transform of
k
.
+ k2 )
[Hint: The quickest way is to use the ”transforms of integrals” theorem.]
F (s) =
s2 (s2
We did an example much like this in class (and in the homework), so the easiest approach is to follow that
example. We first notice that
1
F (s) = 2 L{sin kt}.
s
Using the theorem from class about transforms of integrals, this becomes
(
Z t
t )
1
1
1
1
1
1
= L − cos kt +
F (s) = L
− cos kτ
sin kτ dτ = L
s
s
k
s
k
k
0
0
(
)
t
Z t 1
1
τ
t
1
1
dτ = L
− 2 sin kτ +
.
= L − 2 sin kt +
− cos kτ +
=L
k
k
k
k 0
k
k
0
Therefore,
L−1 {F (s)} = −
7
1
t
sin kt + .
k2
k