Series solution

1. May 17, 2006

JohanL

Im trying to solve the following equation
$$y''(x) + \frac{y'(x)}{x} + \frac{y(x)}{x^2} = 0$$

Then with

$$y(x) = \sum_{i=0}^\infty a_i*x^{i+k}$$

$$y'(x) = \sum_{i=0}^\infty a_i*(i+k)*x^{i+k-1}$$

$$y''(x) = \sum_{i=0}^\infty a_i*(i+k)(i+k-1)*x^{i+k-2}$$

I get

$$\sum_{i=0}^\infty a_i*(i+k)(i+k-1)*x^{i+k-2} + \sum_{i=0}^\infty a_i*(i+k)*x^{i+k-2} + \sum_{i=0}^\infty a_i*x^{i+k-2} = 0$$

but how can i get a recurrence relation from this.
I need something like
$$a_{i+2} = f(i)*a_i$$
But with only the same i+k-2 in all terms i dont know how to proceed.

Last edited: May 17, 2006
2. May 17, 2006

HallsofIvy

Staff Emeritus
Yes, go ahead and write out the formula you get:
$$a_i(i+k)(i+k-1)+ a_i(i+k)+ a_i= a_i((i+k)^2+ 1)= 0$$
Either i+k= 0 or ai= 0 for all i.
The difficulty is that x= 0 is not a "regular" singular point for this equation. That's an "equipotential" (also called "Euler type") equation for which 0 is pretty much the "boundary" between regular singular point and irregular singular point. Multiplying on both sides of the equation by x2 you get x2y"+ xy'+ y= 0. Making the change of variable t= ln x or x= et reduces it to an equation with constant coefficients:
$$\frac{dy}{dx}= \frac{dx}{dt}\frac{dy}{dt}= e^t\frac{dy}{dt}= x\frac{dy}{dt}$$
$$\frac{d^2y}{dx^2}= \frac{d }{dx}(x\frac{dy}{dt})= x^2\frac{d^2y}{dt^2}+ x\frac{dy}{dt}$$
so $$x^2\frac{d^2y}{dx^2}+ x\frac{dy}{dx}+ y= \left(\frac{d^2y}{dt^2}+ \frac{dy}{dt}\right)- \frac{dy}{dt}+ y= \frac{d^2y}{dt^2}+ y= 0$$
Solve that equation for y as a function of t and then replace t by ln x.

3. May 21, 2006

JohanL

Thx, I solved it!