Series Solutions to 2nd Order DEs HELP!

• Nov 23rd 2008, 01:42 PM
Happy Dancer
Series Solutions to 2nd Order DEs HELP!
x^2 y" + 2 x^2 y' -2y = 0

I need to find the indicial equation and find the values of k but I don't understand how to find po and qo.

Thanks
• Nov 23rd 2008, 02:35 PM
shawsend
To calculate the indicial equation, write it as:

$\displaystyle y''+p(x)y'+q(x)y=0$

If it's a regular singular point, $\displaystyle p(x)$ cannot have it its denominator the factor x to a power higher than 1.

Write $\displaystyle p(x)=\frac{p_0}{x}+p_1+p_2x+\cdots$

$\displaystyle q(x)=\frac{q_0}{x^2}+\frac{q_1}{x}+q_2+q_3x+\cdots$

Then it's simple, the indicial equation is:

$\displaystyle c^2+(p_0-1)c+q_0$

In your case, it's $\displaystyle c^2-c-2=0$
• Nov 23rd 2008, 03:38 PM
Happy Dancer
Sorry I still don't understand how you've got that
• Nov 24th 2008, 03:50 AM
shawsend
Ok, your equation is: $\displaystyle x^2y''+2x^2y'-2y=0$ or $\displaystyle y''+2y'-\frac{2}{x^2}y=0$

That means $\displaystyle p(x)=\frac{0}{x}+2+0x+0x^2+\cdots$

and $\displaystyle q(x)=-\frac{2}{x^2}+\frac{0}{x}+0+\cdots$

That is $\displaystyle p_0=0$ and $\displaystyle q_0=-2$. Now just plug it into the indicial equation:

$\displaystyle c^2-c-2=0$
• Nov 24th 2008, 09:58 AM
Moo
Please look here : http://www.mathhelpforum.com/math-he...s-problem.html for a similar problem (with a rough method)
• Nov 25th 2008, 09:17 AM
Happy Dancer
Thanks, shawsend is it the same for q(x), it also cannot have its denominator the factor x to a power higher than 1.