Math 307, Sections A and B, Summer 2009, Solutions to Midterm II
1. A series circuit has a capacitor of 10−5 farad, a resistor of 3 × 102 ohms, and an inductor of 0.2 henry.
The initial charge on the capacitor is 10−6 coulomb and there is no initial current.
(a) Set up an initial value problem modeling this circuit. (2 points)
The initial value problem is
0.2Q00 + 300Q0 + 105 Q = 0,
, Q(0) = 10−6 ,
Q0 (0) = 0
where Q(t) is the charge.
(b) Find the charge on the capacitor and the current through the resistance at any time t.
The characteristic equation has roots
p
−300 ± (300)2 − 4(0.2)105
r=
= −1000 or − 500
0.4
so the solution is of the form
Q(t) = C1 e−1000t + C2 e−500t .
To find the constants C1 and C2 we use the initial conditions:
10−6 = Q(0) = C1 + C2
and the current is
Q0 (t) = −1000C1e−1000t − 500C2 e−500t
so
0 = Q0 (0) = −1000C1 − 500C2
giving us
C1 = −10−6 and C2 = 2 × 10−6
therefor the charge is
Q(t) = −10−6 e−1000t + 2 × 10−6 e−500t .
and the current is
I(t) = Q0 (t) = 10−3 e−1000t − 10−3 e−500t .
1
2. Find the second solution -linearly independent from the first one given- to the second order linear
differential equation
t2 y00 + 2ty0 − 2y = 0
if y1 (t) = t is one solution. Write down the general solution. (8 points)
We have to use the method of reduction of order to find the second solution y2 (t). The equation
rewritten in standard form is
2
2
y00 + y0 − 2 y = 0
t
t
now when we let y2 (t) = v(t)y1 (t) we get
4
v00 + v0 = 0
t
letting u = v0
4
u0 + u = 0
t
using the integrating factor we get
u(t) = e−
so
v(t) =
so
Z
R
4
t dt
= t−4
t−4 dt = −
y2 (t) = −
1
3t3
t
3t3
and the general solution can be written in the form
y(t) = C1 t +
2
C2
.
t2
3. Tell me what you should try for a particular solution for the following non-homogeneous differential
equations. Do not find the particular solution. (2 points each)
(a) y00 + 2y0 + y = 5e−t
The homogeneous equation has solution C1 e−t + C2 te−t so for the particular solution to the
non-homogeneous solution I would try
Y (t) = At2 e−t .
(b) y00 + 2y0 + 5y = sin t
The homogeneous equation has solution C1 e−t cos(2t)+C2 e−t sin(2t) so for the particular solution
to the non-homogeneous solution I would try
Y (t) = A cos t + B sin t
(c) y00 + 2y0 + 5y = 1t
This cannot be done with the method of undetermined coefficients. I would do variation of
parameters. The homogeneous equation has solution C1 e−t cos(2t) + C2 e−t sin(2t) so for the
particular solution to the non-homogeneous solution I would try
Y (t) = u1 e−t cos(2t) + u2 e−t sin(2t)
where u1 (t) and u2 (t) are functions.
√ 21t
+
(d) y00 − 5y + 1 = t2 + t sin 7t + t3 e6t The homogeneous equation has solution C1 e2.5t cos
2
√ 21t
C2 e2.5t sin
For the particular solution to the non-homogeneous solution I would try
2
Y (t) = (At2 + Bt + C) + (t + D)(E sin 7t + F cos 7t) + (Gt3 + Ht2 + It + J)e6t .
3
4. Match the following initial value problems with their solutions and their graphs. Do not try to completely solve them all. Instead, try to characterize the equation and see what the solution should look
like. (2 points each)
(a) y00 + 4y = 0, y(0) = 0, y0 (0) = 1.
Solution : III
Graph: G
y(t) = C1 cos 2t + C2 sin 2t
Periodic, looks like a sine wave.
(b) y00 + 3y0 + 2y = 0, y(0) = 0, y0 (0) = 1.
Solution : VI
y(t) = C1 e−t + C2 e−2t ,
(c) y00 + y0 + 2y = 0, y(0) = 0, y0 (0) = 1.
Graph:B
lim y(t) = 0
t→∞
Solution : V
Graph: H
!
√
√ !
7
7
t + C2 e−t/2 cos
t
2
2
y(t) = C1 e−t/2 sin
We have limt→∞ y(t) = 0. It should look like a sine wave squeezed under an exponential decay.
(d) y00 + 4y = cos t, y(0) = 0, y0 (0) = 0.
Solution : I
Graph: C
y(t) = C1 cos 2t + C2 sin 2t + D cos t + E sin t
This is periodic (repeats itself) but is not like a sine wave since we have a sum of functions with
two different periods.
(e) y00 + 4y = cos 2t, y(0) = 0, y0 (0) = 0.
Solution : VII
Graph: D
y(t) = C1 cos 2t + C2 sin 2t + D1 t cos 2t + D2 t sin 2t
The t term will makes this a wave with growing amplitude. It should look like a wave between
two lines. The values of y(t) will get larger and larger.
(f) y00 + 3y0 + 2y = cos t, y(0) = 0, y0 (0) = 0.
Solution : II
Graph: F
y(t) = C1 e−t + C2 e−2t + D cos t + E sin t
The solution to the homogeneous equation will approach zero. So the function will look like
D cos t + E sin t which is a sine wave as t gets large.
(g) y00 + 0.125y0 + y = cos t, y(0) = 1, y0 (0) = 0.
Solution : IV
Graph: E
!
!
√
√
255
255
−t/16
−t/16
t + C2 e
cos
t + D cos t + E sin t
y(t) = C1 e
sin
16
16
The solution to the homogeneous equation will approach zero. So the function will look like
D cos t + E sin t which is a sine wave as t gets large. You can tell apart the graph of this one and
the previous one from their starting points.
(h) y00 − y0 − 2y = cos t, y(0) = 1, y0 (0) = 0.
Solution : VIII
Graph: A
y(t) = C1 e2t + C2 e−t + D cos(t) + E sin(t)
The exponential factor e2t will dominate when t is large and this will look like e2t for large t.
4
Solutions:
y(t) =
I.
II.
y(t) =
3
1
3
3
sin(t) +
cos(t) − e−2t + e−t
10
10
5
2
y(t) =
III.
IV.
127 −t/16
y(t) = − √
e
sin
255
V.
√
255
t
16
1
sin(2t)
2
!
−t/16
+e
cos
2
y(t) = √ e−t/2 sin
7
√ !
7
t
2
V I.
V II.
V III.
2
1
cos(2t) + cos(t)
3
3
y(t) =
√
255
t
16
!
y(t) = −e−2t + e−t
y(t) = cos(2t) +
t
sin(2t)
4
7 2t 5 −t
3
1
e + e −
cos(t) −
sin(t)
15
6
10
10
5
+ 8 sin(t)
+
)
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