1. ## Power Series Help

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

2. 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?

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

Assume the solution have the form

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

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

and

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

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

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

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

4. 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 :/

5. 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

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

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

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

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

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

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

$\displaystyle y(0)= 5$

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$