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

cheatham (sc36975) – HW04 – um – (53890) 1
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 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 .

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