# Power Series Solutions of ODE's

• Feb 10th 2008, 02:12 AM
nahal
Power Series Solutions of ODE's

consider the differential equation (1+x^2)y''-6y=0.
show that x=0 is an ordinary point of this equation - which i have done.

determine the interval in x for which the equation has a convergent solution of the form y=sigma(akx^k).

Show that the recurrence relation is
ak+2=-[(k-3)/(k+1)]*ak

and hence, determine the first six non-zero terms of the series solution.

• Feb 10th 2008, 09:03 AM
Jhevon
Quote:

Originally Posted by nahal

consider the differential equation (1+x^2)y''-6y=0.
show that x=0 is an ordinary point of this equation - which i have done.

determine the interval in x for which the equation has a convergent solution of the form y=sigma(akx^k).

Show that the recurrence relation is
ak+2=-[(k-3)/(k+1)]*ak

and hence, determine the first six non-zero terms of the series solution.

these problems are a pain to type out with LaTex, so i will give you some references. tell me if they help

see here

and here

and also here

just to start you off. we are expanding around $\displaystyle x_0 = 0$, so we assume a solution of $\displaystyle y = \sum_{k = 0}^{\infty}a_kx^k$

so, $\displaystyle y' = \sum_{k = 1}^{\infty}ka_kx^{k - 1}$

and, $\displaystyle y'' = \sum_{k = 2}^{\infty}k(k - 1)a_k x^{k - 2}$

now your problem is: $\displaystyle (1 + x^2)y'' - 6y = 0$

$\displaystyle \Rightarrow y'' + x^2y'' - 6y = 0$

now plug in the series forms for y and y'' and continue as you see in the links i gave you
• Feb 22nd 2008, 11:52 AM
nahal
Thank you for all your help, it was well appreciated.

There werew some very useful links that were provided.
• Feb 22nd 2008, 11:58 AM
Jhevon
Quote:

Originally Posted by nahal
Thank you for all your help, it was well appreciated.

There werew some very useful links that were provided.

you're welcome.

i suppose you solved the question ok then?