Question

Use power series to solve the differential equation.
y'=xy

Answer 1

I guess the first step is this
\[y'-xy=0\]

Answer 2

hold on ... just let
\[ y = \sum_{n=0}^\infty x^n \]

Answer 3

\[ y = \sum_{n=0}^\infty a_n x^n \]

Answer 4

\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}\]
according to my book, same thing I guess

Answer 5

yes yes ...
\[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0\]
find the recurrence relation.

Answer 6

oh ok
\[y''=\sum_{n=2}^{\infty} n(n-1)c_nx^{n-2}\]
how do I apply these generalizations to this specific problem

Answer 7

hah?? why do you want more trouble? you don't need to find y'' for this.

Answer 8

really, I was just trying to follow this example in the book

Answer 9

I meant: really?

Answer 10

lol .. this example is of second order differential equation. you need it only for second order DE

Answer 11

because there is only a y' that makes it first order?

Answer 12

yep.

Answer 13

ok so I only consider this then, right?
\[y'=\sum_{n=1}^{\infty} nc_nx^{n-1}
\]

Answer 14

yes ... do you have a software called maple?

Answer 15

nope

Answer 16

but before solving for it, what do I sub?

Answer 17

or how do I know what do sub, I think I'm just afraid of sums :S

Answer 18

why do they have (n+2)(n+1) in the example?

Answer 19

from maple i got
y(x) = y(0)+(1/2)*y(0)*x^2+(1/8)*y(0)*x^4+O(x^6)

Answer 20

Honesty I don't like power series method.
here's how you do it. assume the solution has of the form
\[ y = \sum_{n=0}^\infty a_n x^n \]
so we are putting these vales in your DE and hunting a_n's

Answer 21

I'm back...

Answer 22

gtg again...

Answer 23

after putting the value of y in your DE, you get
\[ \sum_{n=0}^\infty n \; c_n x^{n-1} + \sum_{n=0}c_n x^{n+1} = 0 \\
\]
since on the left side you don't have negative power of x, put this \( \sum_{n=0}^\infty n \; c_n x^{n-1} \) = \( \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} \)

Also we can make \( \sum_{n=0}^\infty \; c_n x^{n+1} \) = \(\sum_{n=0}^\infty c_{n-1} x^{n} \)

now we have
\[ \huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + \sum_{n=0}^\infty c_{n-1} x^{n} = 0

\\

\huge \sum_{n=0}^\infty (n+1) \; c_{n+1} x^{n} + c_{n-1} x^{n} = 0 \\
\huge \sum_{n=0}^\infty ((n+1) \; c_{n+1} + c_{n-1} ) x^{n} = 0 \]

Answer 24

since you don't have power's of x on the right side, you have
\[ (n+1) c_{n+1} + c_{n-1} = 0 \\
\text{ or, } c_{n+1} = {- c_{n-1} \over n+1}\]

this is the recurrence relation you get
put n=1, you get

this way you find the coefficients. hence you have solution.