Solutions
Differential Equations - Spring 2012 - Final Exam - Review
1. Find the solution to the initial value problem.
x0 (t) = 2x + y;
y 0 (t) = 8x;
x(0) = 0
y(0) = −6
First, rewrite as a single equation using matrices:
0
x =
2
8
1
0
x
Recall that the general solution will be of the form x = C1 ξ1 er1 t + C2 ξ2 er2 t , where r1 and r2 are the eigenvalues
of the matrix and ξ1 and ξ2 are the corresponding eigenvectors. Thus, the next step will be to find r1 and r2 ,
by finding roots to the characteristic polynomial:
2−r
1
p(r) = det
= (2 − r)(−r) − 8 = r2 − 2r − 8 = (r − 4)(r + 2)
8
0−r
This gives us eigenvalues
r1 = −2 and r2 = 4. To find the corresponding eigenvectors, we row reduce the
augmented matrices A − ri I0 .
r1 = −2
Thus ξ1 =
−1
4
2 − (−2)
8
0
1
0 − (−2) 0
=
4
8
1
2
0
0
=⇒
1
0
1
4
0
0
0
r2 = 4
2−4
1
8
0−4
0
0
=
−2 1
8 −4
0
0
=⇒
1
0
− 21
0
0
0
1
Thus ξ2 =
2
To solve the IVP, we solve the equation
−1
1
0
+ C2
=
4
2
−6
1 0
1 −1 0
1 0
=⇒
=⇒
6 −6
0 1 −1
0 1
x(0) = C1
−1
4
1 0
2 −6
=⇒
−1
0
−1
−1
So C1 = −1 and C2 = −1. The solution is
x(0) = −
−1 −2t
1 4t
e
−
e , or
4
2
x(t) = e−2t − e4t
y(t) = −4e−2t − 2e4t
2. Find the specific solution to the separable differential equation with y(1) = 1; leave your answer in implicit
form.
dy
xy − xy 3 = 10 ln(x)
dx
First, the general solution:
1
dy(y − y 3 ) = 10 ln(x)
x
1
Z
y − y 3 dy = 10
Z
1
ln(x) dx
x
y2
y4
10
−
=
(ln(x))2 + C
2
4
2
2y 2 − y 4 = 20(ln(x))2 + C
Now, the initial conditions:
2 − 1 = 20(ln(1))2 + C = 0 + C, so C = 1
2y 2 − y 4 = 20(ln(x))2 + 1
3. Find the general solution for the linear first-order differential equation.
y 0 + tan(x)y = cos2 (x)
R
First construct the integrating factor µ(x) = e tan(x) dx = eln(sec(x)) = sec(x), and then multiply through to get
the equivalent equation
sec(x)y 0 + sec(x) tan(x)y = sec(x) cos2 (x)
Rewrite the left side as a product rule derivative, and solve for y.
d
[sec(x)y] = cos(x)
dx
Z
sec(x)y = cos(x) dx = sin(x) + C
y=
C
sin(x)
+
= sin(x) cos(x) + C cos(x)
sec(x) sec(x)
4. Find the general solution to the following second order linear non-homogeneous equation. (You may use either
undetermined coefficients or variation of parameters for this problem.)
y 00 − y 0 − 6y = e3t
First, we construct the auxiliary equation in order to find the solution to the homogeneous equation:
r2 − r − 6 = (r − 3)(r + 2) = 0, so the complementary solution is Y (t) = C1 e3t + C2 e−2t
We proceed by undetermined coefficients. In this case, we would assume the particular solution is of the form
f (t) = Ae3t , but because e3t is already represented by the complementary solution (corresponding to “s = 1”),
we must instead use f (t) = Ate3t . Now
f 0 (t) = Ae3t + 3Ate3t = (3At + A)e3t , and
f 00 (t) = 3Ae3t + 3Ae3t + 9Ate3t = (9At + 6A)e3t
To find the coefficients, we solve
f 00 (t) − f 0 (t) − 6f (t) = e3t
((9At + 6A) − (3At + A) − 6At) e3t = 5Ae3t = e3t
Thus, A = 51 , so our general solution is
y(t) =
1 3t
te + C1 e3t + C2 e−2t
5
5. A trust fund is created with an initial deposit of $500,000 and from then on earns interest at an annual rate
of 3%. If the trust beneficiary plans to take a regular salary from the trust to supplement her income, what is
the maximum possible annual salary she can take without causing the principal balance of the trust to shrink?
(Assume that all transactions occur continuously.)
To answer this question, we construct an initial value problem differential equation to model the balance of the
account after t years. Let A(t) be the balance at time t, and S be the annual salary drawn from the account.
Then the equation satisfied by the data in the problem is:
A0 (t) = 0.03A − S,
s(0) = 500, 000
Because this is an autonomous equation, the maximal salary corresponds to the value S which makes A0 (t) = 0.
Thus we solve 0.03A(0) − S = 0 to get S = 0.03 · $500, 000 = $15, 000.
2
6. Find the general solution to the equation
0
x =
−2
1
−1
−2
x
Similar to the above, we first find the roots of the characteristic equation to get the eigenvalues, and use these
to find the eigenvectors. The eigenvalues are r1 = −2 − i and r2 = −2 + i, which means that we can expect the
eigenvectors to also be complex conjugates of each other. Thus, it suffices to find ξ1 :
r1 = −2 − i
−2 − (−2 − i)
−1
0
1
−2 − (−2 − i) 0
=
i
1
−1
i
0
0
=⇒
1 i
0 0
0
0
−i
i
Thus ξ1 =
, and ξ2 =
. A first step toward the general solution will be of the form
1
1
−i (−2−i)t
i (−2+i)t
x(t) = C1
e
+ C2
e
1
1
. . . but to get a real solution, we must finagle the complex solutions as below.
−i −2t
i −2t
e (cos(−t) + i sin(−t)) + C2
e (cos(t) + i sin(t))
1
1
−i
i
= e−2t C1
(cos(t) − i sin(t)) + C2
(cos(t) + i sin(t))
1
1
C1 (−i cos(t) − sin(t)) + C2 (i cos(t) − sin(t))
= e−2t
C1 (cos(t) − i sin(t)) + C2 (cos(t) + i sin(t))
(C2 − C1 )i cos(t) − ((C1 + C2 ) sin(t)
= e−2t
(C1 + C2 ) cos(t) + (C2 − C1 )i sin(t)
x(t) = C1
With new constants D1 := C2 − C1 and D2 := C1 + C2 , this becomes
cos(t) −2t
− sin(t) −2t
x(t) = D1
e
+ D2
e
sin(t)
cos(t)
7. Determine whether the below equation is exact; if it is exact, find the general solution in implicit form.
3x2 + 4y + 7y sin(xy) dx + (4x + 7x sin(xy)) dy = 0
Let M = 3x2 + 4y + 7y sin(xy) and N = 4x + 7x sin(xy). Then to check exactness, we find My and Nx :
My = 4 + 7 sin(xy) + 7xy cos(xy); Nx = 4 + 7 sin(xy) + 7xy cos(xy)
Thus the equations are exact, and we now find a solution template by taking antiderivatives:
Z
ψ(x, y) = 3x2 + 4y + 7y sin(xy) dx = x3 + 4xy − 7 cos(xy) + g(y)
To find g(y), we take ψy :
4x + 7x sin(xy) + g 0 (y)
Thus, g 0 (y) = 0, so our solution is ψ = C, or
x3 + 4xy − 7 cos(xy) = C
8. Find the Laplace transform of the function:

0≤t<1
 0
−1
1
≤ t < π3
f (t) =

π
sin t − 3
t ≥ π3
3
First, we rewrite f as a single function using Heaviside functions:
π
f (t) = −u1 (t) + sin t −
+ 1 u π3 (t)
3
Using the theorem about
Laplace transforms
of step functions, we need to find g(t) so that g(t − 1) = −1, and
h(t) so that h t − π3 = sin t − π3 + 1. The appropriate functions are g(t) = −1 and h(t) = sin(t) + 1. Thus
the Laplace transform of f (t) is
1
1
−π
s
−s −1
3
+e
+
L{f (t)} = e
s
s2 + 1 s
9. Use Laplace transforms to solve the initial value problem:
y 00 − y = sin(t); y(0) = 0, y 0 (0) = 0
Let F (s) = L{y}. Then sF (s) = L{y 0 } and ss F (s) = L{y 00 }. Now we find the Laplace transform of the entire
equation.
1
s2 F (s) − F (s) = 2
s +1
1
1 1
1 1
F (s)(s2 − 1) = 2
= (by some work with partial fractions) −
+
(s + 1)(s2 − 1)
2 s2 + 1 2 s2 − 1
Taking inverse Laplace transforms, we find
1
1
y(t) = − sin(t) + sinh(t)
2
2
10. Use Laplace transforms to solve the initial value problem:
y 00 − 2y 0 − 8y = 0; y(0) = 1, y 0 (0) = 0
Similarly to the above, we suppose that F (s) = L{y}. Then sF (s) − 1 = L{y 0 } and s2 F (s) − s = L{y 00 }, so the
Laplace transform of the equation is
s2 F (s) − s − 2(sF (s) − 1) − 8F (s) = 0
F (s)(s2 − 2s − 8) − s + 2 = 0
F (s) =
s−2
s−2
1 1
2 1
=
=
+
s2 − 2s − 8
(s − 4)(s + 2)
3s−4 3s+2
Finding the inverse transform gives us
y(t) =
1 4t 2 −2t
e + e
3
3
4