Math 2280
Summer 2007
Final exam
You will have two hours to complete this test. You may not use a calculator or other electronic device. Each
problem is worth 10 points. Please be neat so that I can grade your work easily. All solutions should be real-valued.
Unless an initial condition is specified, all solutions should contain the appropriate arbitrary constants.
1. Solve the equation
dy
= 2xy 2 + 3x2 y 2 .
dx
First, we must notice that this problem is seperable. This is more easily done if we rewrite it as
dy
= (2x + 3x2 )y 2 .
dx
We then have that
Z
dy
=
y2
Z
(2x + 3x2 ) dx ⇒ −
1
−1
= x2 + x3 + C ⇒ y = 2
.
y
x + x3 + C
2. Solve the equation
3xy ′ + y = 12x.
Again, we must rewrite this problem in order to better recognize it. If written as
y′ +
1
y = 4,
3x
it is more easily recognized as a first order linear equation. We choose our integrating factor as
Z
1
1
1
dx = exp
ln x = x 3 ,
ρ(x) = exp
3x
3
and multiplying through the equation, we find that
1
1
1 2
x 3 y ′ + x− 3 y = 4x 3 .
3
We also must recognize that the left hand side may be written as
h 1 i
1
Dx x 3 y = 4x 3 .
Integrating both sides gives us that
1
x3 y =
Z
1
4
4x 3 dx = 3x 3 + C.
Solving for y yields
1
y = 3x + Cx− 3 .
3. Given the equation
dx
= (x + 2)(x − 2)2 ,
dt
list the fixed points/equilibrium solutions and determine the stability of each.
The equilibrium solutions satisfy
(x + 2)(x − 2)2 = 0 ⇒ x = +2, +2, −2.
1
20
10
–3
–2
–1
0
1
2
3
4
x
–10
–20
Figure 1: A graph of dx/dt versus x.
To determine their stability, we plot dx/dt as a function of x. This graph should look like Figure 1.
Since dx/dt changes sign from negative to positive at x = −2, it is an unstable fixed point. Because dx/dt does
not change sign at x = 2, it is a semi-stable fixed point.
4. Find the general solution to the equation
y ′′ + 5y ′ + 5y = 0.
To solve this problem, we must first find the roots of the characteristic equation, which is
√
2
25
5
25
5
5
5
2
2
r + 5r + 5 = 0 ⇒ r + 5r +
=
= −5 +
⇒ r+
⇒ r=− ±
.
4
4
2
4
2
2
These roots are real and distinct, so the solution to our equation is
5
y(x) = c1 e− 2 +
√
5
2
5
+ c2 e − 2 −
√
5
2
.
5. Find the general solution to the equation
y ′′ + 4y = 7ex+1 .
Recall that the solution consists of two parts, the complimentary solution yc , which is a solution of the homogeneous equation y ′′ + 4y = 0 , and a particular solution yp to the non-homogeneous problem. To find yc , we first
solve the characteristic equation,
r2 + 4 = 0 ⇒ r = ±2i.
Two complex conjugate roots with zero real part tell us that
yc (x) = c1 cos 2x + c2 sin 2x.
In order to find yp (x), we first guess that it has the form
yp (x) = Aex
for the right choice of A. We then have that
yp′ (x) = yp′′ (x) = Aex ,
2
so that
yp′′ + 4yp = Aex + 4Aex = 5Aex .
For yp to be a solution, we must require that
5A = 7e ⇒ A =
7
e.
5
Therefore, yp (x) = 75 ex+1 , and the general solution is
7
y(x) = c1 cos 2x + c2 sin 2x + ex+1 .
5
6. Find the solution to the initial value problem
−6 2
′
~x =
~x,
−8 4
~x(0) =
1
2
.
As always, we must first find the eigenvalues of the coefficient matrix. These satisfy
−6 − λ
2 0=
= (−6 − λ)(4 − λ) + 16 = −24 + 6λ − 4λ + λ2 + 16 = λ2 + 2λ − 8 = (λ − 2)(λ + 4).
−8
4−λ Thus the eigenvalues are λ1 = 2 and λ2 = −4. We also need the eigenvectors. In order to find and eigenvetor
corresponding to λ1 = 2, we row reduce as follows:
1
−8 2
1 − 14
a
1
∼
=
.
⇒ a = b ~v1 =
−8 2
b
4
0 0
4
To find an eigenvector corresponding to λ2 = −4, we row reduce as follows:
−2 2
1 −1
a
1
∼
⇒ a = b ⇒ ~v2 =
=
.
−8 8
0 0
b
1
The general solution to the system is then
~x(t) = c1 e
2t
1
4
+ c2 e
−4t
1
1
.
The initial conditon implies the augmented matrix equation
1
1
1 1
1
1 1
∼
∼
0
4 1
2
0 −3
−2
0
1
1
3
2
3
.
Therefore, the solution to the initial value problem is
1 2t 1
2 −4t 1
~x(t) = e
+ e
.
4
1
3
3
7. Find the general solution of the system of equations
5 2
′
~x =
~x.
−2 1
As always, we first compute the eigenvalues, which must satisfy
5−λ
2 0=
= (5 − λ)(1 − λ) + 4 = 5 − 5λ − λ + λ2 + 4 = λ2 − 6λ + 9 = (λ − 3)2 .
−2
1−λ 3
Thus we have a repeated eigenvalue λ = 3. Next we seek eigenvectors corresponding to λ = 3. To do so, we row
reduce as follows:
2
2
1 1
−1
∼
⇒ v1 = −v2 ⇒ ~v1 =
.
−2 −2
0 0
1
We need two linearly independent eigenvectors, but this process produced only one. Therefore, the other must
be a generalized eigenvector. To find it, we solve the augmented matrix equation
1 1
1 1
−1
− 12
2
2
−2
.
∼
⇒ v1 = −v2 −
⇒ ~v2 =
0 0
0
1
0
−2 −2
2
The general solution is then
~x(t) = c1 e
3t
−1
1
+ c2 e
3t
− 12
0
−1
1
+t
−1
1
.
8. Find the general solution of the system
′
~x =
The eigenvalues of the coefficient matrix satisfy
1−λ
0 = 3
1
3
~x.
−1 = (1 − λ)2 + 3,
1−λ √
which implies
that λ = 1 ± 3i. We need to find an eigenvector corresponding to one of the eigenvalues. I chose
√
λ = 1 + 3i, and we row reduce to find the eigenvector as follows:
√
√
√ √ √
3
0
−1
− 3i
3i
3
1 − 33 i
~
√
∼
⇒ ~a =
.
⇒ v1 =
, b=
iv2 ⇒ ~v1 =
3
3
0
4
− 3i
3
0
0
Recall that
eλt~v = e(p+qi)t (~a + ~bi) = ept [(cos qt + i sin qt)(~a + ~bi)] = ept [cos qt~a − sin qt~b + i(cos qt~b + sin qt~a)],
so using the real and imaginary parts above as our two linearly independent solutions, we have that
√ √ √
√
√
0
0
3
3
+ c2 et cos 3t
+ sin 2t
.
~x(t) = c1 et cos 3t
− sin 3t
3
3
0
0
9. Find the matrix exponential eAt , where
A=
2
−2
2
−2
.
[Hint: Use the definition of the matrix exponential.]
According to the hint, we should use the definition of the matrix exponential, which is
eAt = I + At +
We have that
A =
2
2
−2 −2
= I + At =
1
0
2
A2 t2
A3 t3
+
+ ···.
2!
3!
2
2
−2 −2
2t
2t
−2t −2t
=
0 0
0 0
=
1 + 2t
2t
−2t 1 − 2t
,
so in actuality,
e
At
0
1
+
4
.
10. For the system
dx
dt
dy
dt
= 2x − 4xy
= −y + 2xy
find all the fixed points, find the linearization of the system about each fixed point, and classify the fixed points
by type and stability of their linearizations.
To find the fixed points, we must solve the system of equations
2x − 4xy = 0
.
−y + 2xy = 0
The first equation implies that either x = 0 or 2 − 4y = 0. The second tells us that either y = 0 or −1 + 2x = 0.
Combining together these possibilities gives us fixed points at (0, 0) and (1/2, 1/2). The Jacobian of the system
is
2 − 4y
−4x
J(x, y) =
.
2y
−1 + 2x
For the fixed point at the origin, the coefficient matrix of the linearization is
2 0
J(0, 0) =
⇒ λ1 = 2, λ2 = −1.
0 −1
This fixed point is a saddle, which is unstable. At (1/2, 1/2), the coefficient matrix of the linearization is
1 1
0 −2
J
=
.
,
1 0
2 2
The eigenvalues of this matrix must satisfy
√
−λ −2
0=
= λ2 + 2 ⇒ λ = ± 2i.
1 −λ
Since the eigenvalues are complex with zero real part, the linearization predicts a stable center for this fixed
point.
11. Use the definition of the Laplace transform to find
L{cos t}.
By definition,
L{cos t} =
Z
∞
e−st cos t dt.
0
If we let u = e−st , du = −se−st dt, dv = cos t dt, and v = sin t, then an integration by parts gives us that
Z ∞
Z ∞
Z ∞
∞
e−st sin t dt (s > 0).
−s sin te−st dt = s
e−st cos t dt = e−st sin t 0 −
If we now let u = e
that
0
0
0
−st
−st
, du = −se
dt, dv = sin t dt, and v = − cos t and integrate by parts again, we find
Z ∞
Z ∞
∞
e−st cos t dt = s −e−st cos t 0 − s
cos t se−st dt
0
⇒ (1 + s2 )
0
Z
∞
0
e−st cos t dt = s ⇒ L{cos t} =
12. Using Laplace transforms, solve the initial value problem
x′ + 2x = t,
5
x(0) = 0.
s
.
1 + s2
First, we transform the equation to get
sX(s) − x(0) + 2X(s) =
1
1
⇒ X(s) = 2
.
s2
s (s + 2)
We then use partial fractions to decompose the right hand side as
1
C
A
B
= + 2+
⇒ 1 = As(s + 2) + B(s + 2) + Cs2
+ 2)
s
s
s+2
s2 (s
= As2 + 2As + Bs + 2B + Cs2 = (A + C)s2 + (2A + B)s + 2B.
⇒ A + C = 0,
2A + B = 0,
2B = 1.
These equations are easily solved to give A = −1/4, B = 1/2, and C = 1/4. Therefore,
X(s) =
−1
1
1
+ 2+
,
4s
2s
4(s + 2)
and consulting the table below, we find that
1
1 1
x(t) = − + t + e2t .
4 2
4
FYI
L{tn } =
n!
sn+1
at
n ∈ Z, n ≥ 0, s > 0
L{e } =
1
s−a ,
6
s>0