Power Series Help

**chalk** · April 4th 2011, 08:32 AM

Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

Thanks

**TheEmptySet** · April 4th 2011, 08:42 AM

Originally Posted by chalk

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Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

Thanks

I am not sure what you tried, but the usual method start like this

Assume the solution have the form

$\displaystyle y=\sum_{n=0}^{\infty}a_nx^n$

Then take some derivatives to get
$\displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1}$

and

$\displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$

Now plug these into the ode and work from here to solve for the $a_n$

**FernandoRevilla** · April 4th 2011, 08:44 AM

Originally Posted by chalk

Hey guys I need help with this Power Series question, I used expl differentiation first then I got stuck in anti differentiation, where do I go after?

After writing $y(x)=\sum_{n=0}^{+\infty}a_nx^n$ , what did you obtain?

Edited : Sorry, I didn't see the previous post.

**chalk** · April 4th 2011, 09:14 AM

Yeah I had those forms and subbed it back into the d.eq

\displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

then

\displaystyle y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

Right?
My equation becomes 2a_2 + a_0 = 0 then to a_2 = -(1/2) a_0

EDIT: Ah I dunno how to get the formula source code up :/

**TheEmptySet** · April 4th 2011, 09:28 AM

Originally Posted by chalk

Yeah I had those forms and subbed it back into the d.eq

\displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

then

\displaystyle y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0

Right?
My equation becomes 2a_2 + a_0 = 0 then to a_2 = -(1/2) a_0

EDIT: Ah I dunno how to get the formula source code up :/

you need to wrap them in math tags
[tex] [ /math] no space between the bracket and forward slash. [ /

If you change your editor to advanced mode you can just click

Tex button in the upper right

$\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}na_nx^{n-1} +\sum_{n=0}^{\infty}a_nx^n = 0$

$\displaystyle \sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \sum_{n=1}^{\infty}na_nx^{n} + \sum_{n=0}^{\infty}a_nx^n = 0$

I think this is what you typed, but you are missing the $x$ and $x^2$ in front of the function and the first derivative

**chisigma** · April 4th 2011, 11:57 AM

Even if 'non reccommended' the most confortable approach is to find y as...

$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (1)

The values of $y(0)$ and $y^{(1)} (0)$ are given by the 'initial conditions'...

$y(0)= 5$

$y^{(1)}(0)= -2$

... and the others are given by the DE...

$y^{(2)} (0) = -x^{2} y^{(1)} (0) + x y(0)= 0$

$y^{(3)} (0) = y(0) -x y^{(1)} (0) -x^{2} y^{(2)}(0)= 5$

$y^{(4)} (0) = -3 x y^{(2)} (0) -x^{2} y^{(3)}(0)= 0$

$y^{(5)} (0) = -3 y^{(2)} (0) -5 x y^{(3)}(0) - x^{2} y^{(4)} (0) = 0$

$y^{(6)} (0) = -8 y^{(3)} (0) -7 x y^{(4)}(0) - x^{2} y^{(5)} (0) = -40$

... and so one, so that the power series is...

$\displaystyle y(x)= 5 -2 x + \frac{5}{6} x^{3} -\frac{1}{18} x^{6} + ...$ (2)

Kind regards

$\chi$ $\sigma$

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