Math 211 Fall 2007: Practice Sketch Solutions for
2nd Midterm
Instructor: Sabin Cautis
Here are sketches of the more complicated parts of the solutions. Beware
that I have not checked the solutions so there may be some errors (you should
use them only for guidance).
1. Find the general solutions of the following problems:
y ′′ + 4y ′ + 8y = 0
y ′′ − 10y ′ + 25y = 0
y ′′ + 6y ′ + 8y = 0
All three ODEs can be solved by the standard method of finding the roots
of the characteristic polynomial. The first has two complex roots −2 ± 2i,
the second has a double root 5 and the third has two real roots −2, −4.
2. Find the Laplace transform of et cos(2t)
We have L(et cos(2t)) = L(cos(2t))(s − 1) =
3. Find the inverse Laplace transform of
We have
3
3
=−
2
s − 2s
2
so the inverse Laplace is
s−1
(s−1)2 +22 .
3
s2 −2s .
1
1
−
s s−2
3 3
− + e2t .
2 2
4. Solve the following system of equations:
x1 + 3x2 + 5x3 = 7
x1 + 5x3 = 1
7x1 + 2x2 = 10
Subtracting the first from second and third gives
x1 + 3x2 + 5x3 = 7
−3x2 = −6
−19x2 − 35x3 = −39
So x2 = 2, x3 = 1/35 and x1 = 6/7.
5. Solve the following initial value problem using the Laplace transform
y ′′ + y ′ + ty = et : y(0) = y ′ (0) = 0
(you may leave some L−1 in your answer).
1
Applying Laplace gives s2 L(y) + sL(y) − L(y)′ = s−1
. We can rewrite
this as
1
L(y)′ − (s2 + s)L(y) =
1−s
which
R 2 is is a linear ODE which we can solve. The integrating factor is
e− s +sds so we get
Z
3
2
3
2
1
L(y) · e−s /3−s /2 =
· e−s /3−s /2 ds
1−s
which just requires a little more simplifying.
1 2 3
−1 5
6. Let A =
and B =
. Compute AB and BA
4 5 −1
0 4
whenever they are defined.
7. Find the general solution of the differential equation: y ′′ +2y ′ +y = 2 cos(t)
Look for a solution yp = A cos(t) + B sin(t). Plug in and solve for A, B.
8. Find the general solution of the differential equation: y ′′ +3y ′ +2y = tan(t)
using the variation of parameters method.
Look for solution yp = ue−t + ve−2t .
yp′ = u′ e−t − ue−t + v ′ e−2t − 2ve−2t
and we set u′ e−t + v ′ e−2t = 0. Then
yp′′ = −u′ e−t + ue−t − 2v ′ e−2t + 4ve−2t
Plugging back and simplifying gives
2u′ e−t + v ′ e−2t = tan(t)
which gives u′ = et tan(t) and v ′ = −e2t tan(t). The rest follows easily.