- #1

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[tex]y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0[/tex]

in form of a power series centered around [tex]x_0=0[/tex].

When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?

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- Thread starter Treadstone 71
- Start date

- #1

- 274

- 0

[tex]y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0[/tex]

in form of a power series centered around [tex]x_0=0[/tex].

When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?

- #2

Physics Monkey

Science Advisor

Homework Helper

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Try multiplying the whole equation by [tex] 1 + x^2 [/tex].

- #3

saltydog

Science Advisor

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Treadstone 71 said:

[tex]y''+\frac{3x}{1+x^2}y'+\frac{1}{1+x^2}y=0[/tex]

in form of a power series centered around [tex]x_0=0[/tex].

When I expanded the y's in power series, I am unable to bring the 1/1+x^2 inside the infinite sum. Can anyone help?

Multiply throughout by [itex](1+x^2)[/itex] first to get:

[tex](1+x^2)y^{''}+3xy^{'}+y=0[/tex]

So wouldn't that first term just be:

[tex]\sum n(n-1)a_n x^{n-2}+\sum n(n-1)a_nx^n[/tex]

right?

- #4

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- #5

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[tex]\sum_{n=0}(n+2)(n+1)a_{n+2}x^{n+2}[/tex]

Are you saying that at this point I can reset the index of x to n and leave a_n the way it is?

- #6

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x^((m+2)-2)=x^m with the coefficients you have. Then subsititute back and let m=n.

- #7

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Ok, got it, I think:

[tex]\sum_{n=0}^{\infty}x^n[a_{n+2}(n+2)(n+1)+a_n[n(n-1)+3n+1]][/tex]

This implies that

[tex]a_{n+2}(n+2)(n+1)+a_n(n+1)^2=0[/tex]

[tex]\sum_{n=0}^{\infty}x^n[a_{n+2}(n+2)(n+1)+a_n[n(n-1)+3n+1]][/tex]

This implies that

[tex]a_{n+2}(n+2)(n+1)+a_n(n+1)^2=0[/tex]

Last edited:

- #8

- 60

- 0

yes, now find your recurence relation and your done.

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