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- May 18th 2008, 06:51 PMszpengchaosolve this Differential equation by indicial equation
- May 18th 2008, 06:55 PMmr fantastic
- May 18th 2008, 07:02 PMszpengchaoproblem
i have problem with the first part

- May 18th 2008, 07:11 PMTheEmptySet
Hint:

$\displaystyle z=\ln(x)$

Use the chain rule

$\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{ x}\frac{dy}{dz}$

Now we need the 2nd derivative using the product and chain rule

$\displaystyle \frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x}\left( \frac{d^2y}{dz^2} \cdot \frac{1}{x}\right)$

from here sub into the equation

Good luck - May 18th 2008, 07:15 PMszpengchaono
no...i mean the 1st question...not the second

- May 18th 2008, 07:23 PMTheEmptySet
- May 18th 2008, 07:24 PMszpengchaoyeah
yeah.... i have problem with choosing point for this particular question...

would you like to post ur steps? - May 18th 2008, 07:45 PMszpengchaochoosing ordinary point
well, x=0 is a regular singular point, right?

then , i set y= sigma( A_n * x^(n+b)) into the equation. then i got something like:

sigma( A_n * (n^2+1) = 0) well, this means A_n = 0 for all n...then how can i continue? - May 18th 2008, 08:34 PMTheEmptySet
$\displaystyle y=\sum_{n=0}^{\infty}c_nx^{n+r} $

$\displaystyle y'=\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1} $

$\displaystyle y''=\sum_{n=2}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2} $

$\displaystyle x^2y''+xy'+y=0$

$\displaystyle x^2\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}+x\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$

$\displaystyle \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r}+\sum_{n=0}^{\infty}c_n(n+r)x^{n+r}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$

$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)(n+r-1)+(n+r)+ 1 \right)=0 $

$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)[(n+r-1)+1]+ 1 \right)=0 $

$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0 $

I think this is where you are stuck. - May 18th 2008, 08:36 PMszpengchaoyup
yup...then? how can that be zero?

- May 18th 2008, 08:49 PMTheEmptySet
- May 18th 2008, 10:13 PMmr fantastic
- May 19th 2008, 01:21 AMmr fantastic
- May 19th 2008, 08:56 AMTheEmptySet
Thanks Mr. Fantastic,

I forgot that $\displaystyle c_0 \ne 0$ in this method so

$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0 \iff c_0x^{r}(r^2+1)+c_nx^{n+r}\left(\sum_{n=1}^{\infty }(n+r)^2+ 1 \right)=0$

so $\displaystyle c_n=0, \forall n \ge 1$

$\displaystyle r^2=1 \iff r=\pm i$

Then the rest follows.

Thanks again!(Clapping)