Solve ODE by Power Series - Singular Point

Hi all,

So I am just wondering if anyone can confirm that the following is a solution of the DE. I'm not entirely sure how to differentiate a factorial so I am at crossroads for checking to see if it satisfies the DE. This was a question off an exam I wrote this morning, and I'm just wondering if that solution is indeed correct.

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

Solution = $\displaystyle 1 + \sum^{\infty}_{n=1}{\frac{x^{3n}}{(n^2)!}$

Thanks in advance! :)

Re: Solve ODE by Power Series - Singular Point

y=a+b*x+c*x^2+d*x^3+e*x^4...

y'=b+2c*x+3d*x^2+4e*x^3 and so on ...you equate multipliers in both sides of DE...

Re: Solve ODE by Power Series - Singular Point

Quote:

Originally Posted by

**sjmiller** Hi all,

So I am just wondering if anyone can confirm that the following is a solution of the DE. I'm not entirely sure how to differentiate a factorial so I am at crossroads for checking to see if it satisfies the DE. This was a question off an exam I wrote this morning, and I'm just wondering if that solution is indeed correct.

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

Solution = $\displaystyle 1 + \sum^{\infty}_{n=1}{\frac{x^{3n}}{(n^2)!}$

Thanks in advance! :)

You don't need to differentiate the factorial, it's a constant. x is the variable.