Solving Equation: y''(x) + \frac{y'(x)}{x} + \frac{y(x)}{x^2} = 0

[JohanL]

Im trying to solve the following equation
<br /> y&#039;&#039;(x) + \frac{y&#039;(x)}{x} + \frac{y(x)}{x^2} = 0<br />

Then with

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

y&#039;(x) = \sum_{i=0}^\infty a_i*(i+k)*x^{i+k-1}

y&#039;&#039;(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 don't know how to proceed.

 

[HallsofIvy]

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 a_i= 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= e^t 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.

 

[JohanL]

Thx, I solved it!