- #1

Gogeta007

- 23

- 0

## Homework Statement

y'' + x

^{2}y' + xy = 0

## Homework Equations

using power series:

y'' = [tex]\Sigma[/tex]c

_{n}n(n-1)x

^{n-2}(n = 2 -> infinity)

y' = [tex]\Sigma[/tex]c

_{n}(n-1)x

^{n-1}(n = 1 -> infinity)

y = [tex]\Sigma[/tex]c

_{n}x

^{n}(n = 0 -> infinty)

## The Attempt at a Solution

by setting the above equation with its corresponding sumation formulas I get the following:

[tex]\Sigma[/tex]c

_{n}n(n-1)x

^{n-2}+ [tex]\Sigma[/tex]c

_{n}(n-1)x

^{n+1}+ [tex]\Sigma[/tex]c

_{n}x

^{n+1}

to make all the x's start at the same power I pull out 2 terms from the first sumation and 1 term from the last eq. so all x's start at the power of 2.

this yields:

2c

_{2}+ 6c

_{3}x + [tex]\Sigma[/tex]

_{n=4}c

_{n}n(n-1)x

^{n-2}+ [tex]\Sigma[/tex]c

_{n}(n-1)x

^{n+1}+ c

_{0}x + [tex]\Sigma[/tex]

_{n=1}c

_{n}x

^{n+1}

on the first sumation k=n-2, so n= k+2, on the second sumation k=n+1 and n=k-1 and the same for the third sumation.

substituting k's in their respective sumations we get:

c

_{0}x+2c

_{2}+6c

_{3}x + [tex]\Sigma[/tex]

_{k=2}x

^{k}{ c

_{k+2}(k+2)(k+1) + c

_{k-1}(k-1) = c

_{k-1}

from the last two terms we can factor a c

_{k-1}and cancel out the 1's left leaving us only with kc

_{k-1}

===========

from the identity that the coefficients should add up to zero, we conclude that c

_{2}=0 and c

_{3}= -c

_{0}/6

for the sumation part, I solve for c_k+2 = -{c

_{k-1}(k)}/(k+2)(k+1)

and plugging in values of k (1,2,3,4,5,6,7,8,9,10) I can see that k=3 = 0, k=6 = 0, k=9 = 0

for the other values I get sumations with c

_{0}and sumations with c

_{1}

Im writing them the way I saw the teacher writing them (since the book does without the multiplication symbol (big pi))

I get the following:

c

_{0}[tex]\Sigma[/tex]

_{n=??}{ (-1)

^{n}[tex]\Pi[/tex]

_{m=0}

^{n-1}(3m+1) } / 12*11*9*8*6*5*3*2

where big pi starts at m =0 and ends at m = n-1

for c_0

and for c_1

c

_{1}[tex]\Sigma[/tex]

_{n=??}{ (-1)

^{n}[tex]\Pi[/tex]

_{m=0}

^{n-1}(3m+2) } / 10*9*7*6*4*3

I couldnt find a pattern for the denominators. . . im thinking its two patterns ( for even and odds). . .anyways I'll do that later, my question is the following:

How do I know where to start the sumations of sigma in the answer, you know n=something ?? (why?)

in the book it says that the c

_{0}sumation has the x

^{3},x

^{6},x

^{9}. . . and that c

_{1}sumation has the x

^{4},x

^{7},x

^{10}

how do I know which sumation has what powers of x?

Some of the answers (in class) start with 1 + [tex]\Sigma[/tex]. . . or with x + [tex]\Sigma[/tex]. . .

how do I know when to pull out the first term? and again, on what n to start for the sumation?

Thanks a lot for reading trough this and thanks in advance for your help.