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Differential equations and linear algebra, midterm exam 2 answer key,
spring term 2014.
Instructor: A. Salch
Some instructions:
• Show your work on all problems. You need to include at least enough detail
to make it clear to the grader what your line of reasoning is, in arriving
at your answer, for each problem on the exam. There is the possibility
of receiving partial credit on problems where you have not arrived at the
correct answer but made some progress toward the correct answer.
• You can write with a pencil or a pen of any color other than red. You may
not use any of the following things: computers, calculators, phones, iPods
or other digital devices, or any human being other than yourself. You can
use any books or notes you have with you.
• You probably don’t need to be reminded of this: don’t cheat. If you are
caught cheating, you will (for starters) receive no credit on this exam, and
you will have to go through the university’s disciplinary process, including
an expulsion hearing. This is a lot worse than failing this exam or even
failing this entire class.
For graders’ use:
Problem number Points received
Points possible
1
20
2
20
3
20
Total
60
1
2
Problem 1. Solve the initial value problem:
y 00 + 3y 0 + 2y
=
sin t
y(0)
=
0
=
0.
0
y (0)
(Hint: you can do this one with either the method of undetermined coefficients or
with variation of parameters—either method will work fine.)
Solution: First we solve the homogeneous equation y 00 + 3y 0 + 2y = 0. The
auxiliary equation is r2 + 3r + 2 = 0, and the quadratic formula easily gives its
roots, r = −1 and r = −2. So the solutions to the homogeneous equation are
y = c1 e−t + c2 e−2t , for all constants c1 , c2 .
Now we use the method of undetermined coefficients to find one solution to the
non-homogeneous ODE y 00 + 3y 0 + 2y = sin t. Let y = A sin t + B cos t. Then
y 0 = A cos t − B sin t and y 00 = −A sin t − B cos t, so
sin t
= y 00 + 3y 0 + 2y
=
(sin t)(A − 3B) + (cos t)(B + 3A), and
A − 3B
=
1,
B + 3A
=
0.
Easy arithmetic then yields A =
non-homogeneous ODE is
1
10
and B =
y = c1 e−t + c2 e−2t +
−3
10 .
So the general solution to our
−3
1
sin t +
cos t.
10
10
3
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4
Problem 2. Find the general solution to the differential equation:
t2 y 00 + ty 0 + y = t2 .
Solution: First divide through by t2 so that the leading coefficient is 1:
1
1
y 00 + y 0 + 2 y = 1.
t
t
Now the homogeneous Cauchy-Euler equation y 00 + 1t y 0 + t12 y = 0 has characteristic
equation r2 + 1 = 0, whose roots are r = ±i, so two linearly independent solutions
to y 00 + 1t y 0 + t12 y = 0 are y1 = sin ln t and y2 = cos ln t. Their Wronskian is
W
1
1
(sin ln t)(− sin ln t) − (cos ln t)( cos ln t)
t
t
−1
=
.
t
=
Now we use variation of parameters: our general solution to y 00 + 1t y 0 +
y = y1 v1 + y2 v2 , where
Z
− cos ln t
v1 =
dt
−1
t
Z
=
t cos ln t dt
Z
=
e2u cos u du,
1
t2 y
= 1 is
making the substitution u = ln t (and observing that t2 = e2u ). Similarly,
Z
sin ln t
dt
v2 =
−1
t
Z
=
−t sin ln t dt
Z
=
−e2u sin u du.
(Getting at least this far will get you most of the credit for this problem.) Now we
use integration by parts to compute these integrals:
Z
Z
e2u sin u du = −e2u cos u + 2 e2u cos u du
Z
2u
2u
= −e cos u + 2e sin u − 4 e2u sin u du, so
Z
2
−1 2u
e cos u + e2u sin u + c1
e2u sin u du =
5
5
Z
Z
2u
2u
e cos u du = e sin u − 2 e2u sin u du
=
2 2u
1
e cos u + e2u sin u + c2 .
5
5
5
Putting it all together:
y
= y1 v1 + y2 v2
1 2 2
2
=
t sin ln t + t2 sin ln t cos ln t + c1 sin ln t
5
5
1 2
−2 2
2
+ t cos ln t +
t sin ln t cos ln t + c2 cos ln t
5
5
1 2
t + c1 sin ln t + c2 cos ln t.
=
5
6
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7
Problem 3. Find all functions f such that f 000 = f .
(Hint: write this as a third-order differential equation, turn it into a system of
three first-order differential equations, and use matrix methods to solve it.)
Solution: The equation y 000 +y = 0 can be written as three first-order equations:
x1
= y
x2
= x01
x3
= x02 ,
x01
= x2
x02
= x3
x03
= x1 ,
yielding equations
and hence the matrix

0




M =
0




1
0
0




1
.



1
0
0
3
Setting det(M −
λid)
=
0
yields
the
polynomial
−λ
+ 1 = 0, whose roots are λ = 1
√
3
−1
and λ = 2 ± 2 i. Computing the associated eigenvectors yields yields:
 
1
 
 
 
 

λ = 1 : c1 
1
 
 
 
1

λ=
√
−1
3
±
2
2






: c2 





1
−1
2
−1
2
and hence












 + i










√
0
√
− 3
2
√
3
2












√
−1
3
3
t + c4 e 2 t cos
t.
2
2
(There are a lot of different but equivalent ways of writing the final answer, and
of course any of them are√fine; the essential
thing is that y can be any linear
√
−1
−1
3
3
t
t
t
2
2
combination of e , e
sin 2 t, and e
cos 2 t.
x1 (t) = y(t) = c1 et + c3 e
−1
2 t
sin
8
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