# Find the general solution?

1. Jul 29, 2013

### Success

1. The problem statement, all variables and given/known data
Find the general solution of (x+1)2y"+3(x+1)y'+0.75y=0 that is valid in any interval not including the singular point.

2. Relevant equations
y=xr
y'=rxr-1
y"=r(r-1)xr-2
(x+1)2(r(r-1)xr-2)+3(x+1)(rxr-1)+0.75xr=0
3. The attempt at a solution
What to do next?

2. Jul 29, 2013

### Zondrina

This looks like a good candidate for a power series solution. Do you know what an ordinary point is?

3. Jul 29, 2013

### Success

4. Jul 30, 2013

### ehild

You are on the right track, but change the variable to t=x+1, and find the solution in the form y=tr (y=(x+1)r).

ehild

5. Jul 31, 2013

### epenguin

I think this can be solved analytically.

It is not indispensible, but you could make the variable X = (x + 1) to have a slightly tamer looking equation, as d/dx = d/dX OK? Edit: I see already suggested.

Then you might recognise this as the linear homogeneous equation. Bearing out what I said in https://www.physicsforums.com/showpost.php?p=4458709&postcount=5 I easily found in Piaggio Art. 40 how to treat this kind. I don't say your substitutions won't work too, but Piaggio gives substitute X = et. You are in the end able to express in t without X and get a linear d.e.

I get
$$4 \frac{d^2y}{dt^2} + 16\frac{dy}{dt} + 3y = 0$$
but don't rely on me, maybe that should be 8dy/dt after all*, it will be something solvable anyway.

Offhand I don't see what the 'singular point' is about, is this X = y = 0?

I would be glad to see the solution results here and what the s.p. is about.

*Edit: Gives nice factorisation! It must be that!

Last edited: Jul 31, 2013
6. Aug 1, 2013

### ehild

The solutions can be found in the form y=(x+1)r. Substituting back, r1=-1/2 and r2=-3/2. So the singular point is x=-1.

ehild