Quiz #1
Math 20D
January 16
SOLUTIONS
1. (§2.1, #30. 10 points) Find the value of y0 for which the solution of the initial value problem
y 0 − y = 1 + 3 sin t,
y (0) = y0
remains finite as t → ∞.
ANSWER:
(1) p (t) = −1
R
(2) µ (t) = e p(t)dt = e−t (some students will be able to figure this out directly, that is OK)
(3) Thus
µZ
¶
1
y (t) =
g (t) µ (t) dt + C
µ (t)
µZ
¶
t
−t
=e
e (1 + 3 sin t) dt + C
µ
¶
Z
t
−t
−t
= e −e + 3 e sin tdt + C
(Give say 7 points
R if students get to here correctly.)
(4) Compute e−t sin tdt by integration by parts twice. In particular: Let
Then
u = sin t,
dv = e−t dt,
du = cos tdt
v = −e−t .
Z
Z
e−t sin tdt = −e−t sin t +
e−t cos tdt.
Now let
dv = e−t dt,
u = cos t,
v = −e−t .
du = − sin tdt
Then
Z
Z
e−t cos tdt = −e−t cos t −
Hence
Z
Z
e
so that
e−t sin tdt.
−t
sin tdt = −e
Z
−t
sin t − e
−t
cos t −
e−t sin tdt,
1
e−t sin tdt = − e−t (sin t + cos t) .
2
So
µ
y (t) = e
t
−e
= −1 −
+3
e
−t
sin tdt + C
3
(sin t + cos t) + Cet .
2
Now y (0) = y0 , so that
y0 = y (0) = −1 −
Therefore
¶
Z
−t
3
5
+ C = − + C.
2
2
µ
¶
5 t
3
e.
y (t) = −1 − (sin t + cos t) + y0 +
2
2
The only way this can be bound is if
y0 +
that is
5
= 0,
2
5
y0 = − .
2
2. (§2.2, #24. 10 points) Solve the initial value problem
y0 =
2 − ex
,
3 + 2y
y (0) = 0
and determine where the solution attains its maximum value.
ANSWER: This is a separable equation:
(3 + 2y) dy = (2 − ex ) dx,
so (integrate)
3y + y 2 = 2x − ex + C.
Since y (0) = 0, we have
0 = −1 + C,
so that C = 1. We get
2x − ex + 1 = 3y + y 2 =
so that
µ
3
y+
2
µ
¶2
3
9
y+
− ,
2
4
¶2
= 2x − ex +
13
.
4
We take the positive square root to get:
3
y=− +
2
r
2x − ex +
13
.
4
The x value where y attains its maximum is where
2x − ex +
13
4
attains its maximum. Settin the derivative to be zero, we get
µ
¶
d
13
x
0=
2x − e +
dx
4
= 2 − ex .
So it is where ex = 2, that is,
x = ln 2.