How to Solve x^2Y''(x) + xY'(x) - CY(x)=0 ODE?

[sara_87]

Homework Statement

Find the general solution of the following ODE:
x^{2}Y''(x) + xY'(x) - CY(x)=0

where C is a constant.

Homework Equations

The Attempt at a Solution

First I want to do this in the case where C=0; this gives:
x^{2}Y''(x) + xR'(x)=0

How do i solve this ODE? Any hints? Thank you.

 

[HallsofIvy]

Please show some work! Do you know anything about "Euler type" equations? (Also called "equipotential" equations.)

 

[sara_87]

I didn't show any working because i got stuck at that point :)
I know how to find the general solution of ODEs but not when there are functions of x infront of the Y'' and Y'.
I don't know about Euler type Equations.

Thank you

 

[gabbagabbahey]

Have you learned the method of power series solutions yet?

 

[sara_87]

no.
This question is in the section of partial differential equation.
I have the two equations:
xY''(x) + xY'(x) - CY(x)=0
and
ZZ''(t)+CZ(t)=0
I can do the second one since there's no function of t infron of the Z. But the first one...I haven't come across something like that.

 

[gabbagabbahey]

You MUST have been taught at least one method of solving such ODE's...what methods have you learned?

 

[sara_87]

the lecturer gave us an example. He found the general seperable solution of Laplace's Equation: V_{xx}+V_{yy}=0

Where the PDE was converted into the ODEs:
X''(x)-\lambdaX(x)=0
Y''(y)+\lambdaY(y)=0

This is simple to solve.

My original question was: find the general solution of:

V_{xx}+(1/x)V_{x}+(1/x^2)V_{tt}
then i said: let V(x,t)=Y(x)Z(t)
so:
V_{xx}=Z(t)Y''(x)
V_{tt}=Y(x)Z''(t)
I substituted these into the original:
V_{xx}+(1/x)V_{x}+(1/x^2)V_{tt}
and divided by Z(t)Y(x) and then multiplied by x^2; to get:
x^2Y''(x)+xY'(x)-CY(x)=0
Z''(t)+CZ(t)=0

 

[sara_87]

sorry, the powers are supposed to be subscripts.