M408D SPRING 2015 HOMEWORK #4 (1) - cheatham (sc36975) HW04...

cheatham (sc36975) – HW04 – um – (53890) 1
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Multiple-choice questions may continue on
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before answering.
001 10.0 points
Find all nonzero values of k for which the
function y = A sin kt + B cos kt satisfies the
di ff erential equation
y
′′
+ 25 y = 0
for all values of A and B .
1. k = 25
2. k = − 5
3. k = − 25
4. k = 25 , − 25
5. k = 5
6. k = 5 , − 5 correct
Explanation:
We will begin by solving for y
′′
.
y = A sin kt + B cos kt
y
′
= Ak cos kt − Bk sin kt
y
′′
= − Ak
2
sin kt − Bk
2
cos kt
= − k
2
( A sin kt + B cos kt )
= − k
2
y.
We compute
y
′′
+ 25 y = − k
2
y + 25 y
= (25 − k
2
) y.
Hence y
′′
+25 y = 0 if and only if k
2
= 25, i.e.,
if and only if k = ± 5. Hence,
k = 5 , − 5 .
002 10.0 points
Find all values of r for which the function
y = e
rt
satisfies the di ff erential equation
y
′′
− 2 y
′
− 8 y = 0 .
1. r = − 4 , 2
2. r = − 2 , 4 correct
3. r = − 8 , − 2
4. r = 2 , 8
5. r = − 2
6. r = 8
Explanation:
We will begin by solving for y
′
and y
′′
.
y = e
rt
y
′
= re
rt
y
′′
= r
2
e
rt
.
We compute
y
′′
− 2 y
′
− 8 y = r
2
e
rt
− 2 re
rt
− 8 e
rt
= ( r
2
− 2 r − 8) e
rt
= ( r + 2)( r − 4) e
rt
.
Hence y
′′
− 2 y
′
− 8 y = 0 if and only if
k = − 2 , 4. Thus,
k = − 2 , 4 .
003 10.0 points
The family of solutions to the di ff erential
equation y
′
= 10 xy is y = Ce
5 x 2
.
Find the solution that satisfies the initial
condition y (3) = 2.
1. y = 2 e
5( x
2
+9)
2. y = 2 e
5( x
2
− 9)
correct

cheatham (sc36975) – HW04 – um – (53890) 2
3. y = 2 e
− 5( x
2
− 9)
4. y = 2 e
5( x − 3)
2
5. y = e
5( x
2
− 9)
+ 1
Explanation:
This equation contains only one unknown
constant, so we will substitute the indicated
values of x and y into the equation to solve
for C as follows:
y (3) = Ce
5(3)
2
= Ce
45
.
This shows that y (3) = 2 if and only if
C = 2 e
− 45
. Thus,
y ( x ) = 2 e
5 x 2
e
− 45
.
Hence,
y ( x ) = 2 e
5( x
2
− 9)
.
004 10.0 points
The family of solutions to the di ff erential
equation y
′
= − 3 y
2
is y =
1
3 x + C
.
Find the solution that satisfies the initial
condition y ( − 2) = − 5.
1. y ( x ) =
5
3( x + 2) − 1
2. y ( x ) =
5
15( x + 2) − 1
correct
3. y ( x ) =
1
3( x + 2) − 5
4. y ( x ) =
25
3( x + 2) − 5
5. y ( x ) =
1
15( x + 2) − 5
Explanation:
This equation contains only one unknown
constant, so we will substitute the indicated
values of x and y into the equation to solve
for C as follows:
y ( − 2) =
1
3( − 2) + C
=
1
C − 6
.
This shows that y ( − 2) = − 5 if and only if
C = −
1
5
+ 6. Thus
y ( x ) =
1
3 x −
1
5
+ 6
=
1
3( x + 2) −
1
5
.
Hence,
y ( x ) =
5
15( x