Mat 303 Calculus Iv Homework 7 Worksheet With Answers - State University Of New York At Stony Brook - 2013 Page 2

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MAT 303 Spring 2013
Calculus IV with Applications
x/3
We use the product rule to compute its derivative; after factoring out e
, this is
x
x
x
x
1
1
1
1
x/3
x/3
y
c
e
cos
sin
c
e
sin
cos
.
1
2
3
3
3
3
3
3
3
3
Evaluating at x
0 and applying the initial conditions, and noting that the sin terms
vanish,
c
c
1
2
y 0
c
3,
y 0
4.
1
3
3
Then c
3, and c
3 4
c
/3
5 3, so the solution to the IVP is
1
2
1
x
x
x/3
x/3
y
3e
cos
5 3e
sin
.
3
3
3
0, y 0
1, y 0
1, y 0
3.3.24. Solve the IVP 2y
3y
2y
3.
3
2
Solution: The characteristic equation for this DE is 2r
3r
2r
0, which factors as
2
r 2r
r 2r
1 r
3r
2
2
0.
Then r
0, r
1/2, and r
2 are its (distinct real) roots, so the general solution is
x/2
2x
y
c
c
e
c
e
.
1
2
3
Its first and second derivatives are
1
1
x/2
2x
x/2
2x
y
c
e
2c
e
,
y
c
e
4c
e
.
2
3
2
3
2
4
Evaluating at x
0 and applying the initial conditions, we have the linear system
1
1
c
c
c
1,
c
2c
1,
c
4c
3.
2
3
2
3
2
3
1
2
4
1
From the third equation, c
12
16c
. Substituting this into the second,
12
2
3
2
16c
2c
1, so 10c
5, and c
1/2. Then c
12
8
4, so c
1
c
c
3
3
3
3
2
2
3
1
7
x/2
1
2x
7/2. Thus, the solution to the IVP is y
4e
e
.
2
2
3
2x/3
0 is y
e
3.3.34. One solution to the DE 3y
2y
12y
8y
. Find the general
solution.
3
2
Solution: The characteristic equation for the DE is 3r
2r
12r
8
0. Since r
2/3
is a root, we expect r
2/3 to be a linear factor of this polynomial. In fact, 3 r
2/3
3r
2 is seen to be a linear factor, so that this polynomial factors as
2
3r
2 r
4
0.
2
Thus, the two other roots are r
2i, from the r
4 factor, so the general solution is
2x/3
c
e
c
cos 2x
c
sin 2x.
1
2
3
2

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