solve this Differential equation by indicial equation

**mr fantastic** · May 18th 2008, 06:55 PM

Originally Posted by szpengchao

Given the form of the DE, I would assume that the first part requires you to find a solution of the form $Ax^n$ .

Where are you stuck in the second part?

**szpengchao** · May 18th 2008, 07:02 PM

i have problem with the first part

**TheEmptySet** · May 18th 2008, 07:11 PM

Originally Posted by szpengchao

Hint:
$z=\ln(x)$

Use the chain rule

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

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

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

from here sub into the equation

Good luck

**szpengchao** · May 18th 2008, 07:15 PM

no...i mean the 1st question...not the second

**TheEmptySet** · May 18th 2008, 07:23 PM

Originally Posted by szpengchao

no...i mean the 1st question...not the second

Are you talking about using the method of frobenius to obtain series solutions to the ODE?

**szpengchao** · May 18th 2008, 07:24 PM

yeah.... i have problem with choosing point for this particular question...
would you like to post ur steps?

**szpengchao** · May 18th 2008, 07:45 PM

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?

**TheEmptySet** · May 18th 2008, 08:34 PM

Originally Posted by szpengchao

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?

$y=\sum_{n=0}^{\infty}c_nx^{n+r}$

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

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

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

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

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

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

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

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

I think this is where you are stuck.

**szpengchao** · May 18th 2008, 08:36 PM

yup...then? how can that be zero?

**TheEmptySet** · May 18th 2008, 08:49 PM

Originally Posted by szpengchao

yup...then? how can that be zero?

I don't know besides the trivial solution. Maybe someone can find my mistake.

**mr fantastic** · May 18th 2008, 10:13 PM

Originally Posted by TheEmptySet

I don't know besides the trivial solution. Maybe someone can find my mistake.

Using the substitution given in the second part, I get $y = A \sin (\ln x) + B \cos (\ln x)$ ......

**mr fantastic** · May 19th 2008, 01:21 AM

Originally Posted by mr fantastic

Given the form of the DE, I would assume that the first part requires you to find a solution of the form $Ax^n$ .

Where are you stuck in the second part?

Do this and you get $n^2 + 1 = 0 \Rightarrow n = \pm i$ .

Therefore

$y = A_1 x^i + A_2 x^{-i} = A_1 e^{i \ln x} + A_2 e^{-i \ln x} = B_1 \cos (\ln x) + B_2 \sin (\ln x)$

by the magic of Euler's equation and the usual identity.

**TheEmptySet** · May 19th 2008, 08:56 AM

Thanks Mr. Fantastic,

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

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

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

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

Then the rest follows.

Thanks again!

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