# Homework Help: ODE: find general solution

1. Mar 8, 2009

### sara_87

1. The problem statement, all variables and given/known data

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

where C is a constant.

2. Relevant equations

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 8, 2009

### HallsofIvy

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

3. Mar 8, 2009

### 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

4. Mar 8, 2009

### gabbagabbahey

Have you learned the method of power series solutions yet?

5. Mar 8, 2009

### 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 havent come across something like that.

6. Mar 8, 2009

### gabbagabbahey

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

7. Mar 8, 2009

### 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)-$$\lambda$$X(x)=0
Y''(y)+$$\lambda$$Y(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

8. Mar 8, 2009

### sara_87

sorry, the powers are supposed to be subscripts.