Math 2280 - Assignment 11 - University of Utah

Math 2280 - Assignment 11

Dylan Zwick Fall 2013

Section 8.1 - 2, 8, 13, 21, 25 Section 8.2 - 1, 7, 14, 17, 32 Section 8.3 - 1, 8, 15, 18, 24

1

Section 8.1 - Introduction and Review of Power Series

8.1.2 - Find the power series solution to the differential equation

y = 4y,

and determine the radius of convergence for the series. Also, identify the series solution in terms of familiar elementary functions.

Solution - We set up the power series

y(x) = cnxn,

n=0

y(x) = ncnxn-1.

n=1

If we plug these into our differential equation we get:

ncnxn-1 - 4 cnxn = 0

n=1

n=0

[(n + 1)cn+1 - 4cn]xn = 0.

n=0

Using the identity principle from this we get the recursion relation:

cn+1

=

4cn n+1

.

2

The first few terms are

c0 = c0,

c1

=

4c0 1

,

c2

=

4c1 2

=

42c0 2?1

,

c3

=

4c2 3

=

3

43c0 ?2?

1

,

and in general,

cn

=

4nc0 n!

.

The radius of convergence for our series is

lim

n

cn cn+1

4n

= lim n

n! 4n+1

=

lim

n

n

+ 4

1

=

.

(n+1)!

So, the power series converges for all x, and the solution to this differential equation is:

y(x) = c0

4nxn n!

=

c0

(4x)n n!

=

c0e4x.

n=0

n=0

But, we already knew that, didn't we! :)

3

8.1.8 - Find the power series solution to the differential equation

2(x + 1)y = y,

and determine the radius of convergence for the series. Also, identify the series solution in terms of familiar elementary functions.

Solution - We set up the power series

y(x) = cnxn,

n=0

y(x) = ncnxn-1.

n=0

If we plug these into our differential equation we get:

2ncnxn + 2ncnxn-1 - cnxn = 0

n=1

n=1

n=0

2ncnxn + (2(n + 1)cn+1 - cn)xn = 0.

n=1

n=0

Using the identity principle the x0 term gives us:

2c1

-

c0

=

0

c1

=

c0 2

.

The higher order terms give us:

4

cn+1

=

(1 - 2(n

2n)cn + 1)

.

The first few terms are

c0 = c0,

c1

=

c0 2

,

c2

=

-

c1 2?2

=

-

c0 22 ? 2!

,

c3

=

-

3c2 2?3

=

3c0 23 ? 3!

,

c4

=

-

5c3 2(4)

=

-

15c0 24 ? 4!

,

and in general,

cn

=

(-1)n+1(2n - 2nn!

3)!!c0 ,

for n 2. Here (2n - 3)!! means the product of all the odd integers up to (2n - 3). The radius of convergence for our series is

lim

n

cn cn+1

c0 (2n-3)!!(-1)n+1

=

lim

n

2n n! c0 (2n-1)!!(-1)n+2

=

lim

n

2(n 2n

+ -

1) 1

=

1.

2n+1 (n+1)!

So, the power series converges for all |x| < 1, and the solution to this differential equation is:

y(x) = c0

1

+

x 2

+

(-1)n+1(2n 2nn!

-

3)!!

xn

.

n=2

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