What Is the General Solution of the Differential Equation Involving (x+1)?

[Success]

Homework Statement

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.

Homework Equations

y=x^r
y'=rx^r-1
y"=r(r-1)x^r-2
(x+1)²(r(r-1)x^r-2)+3(x+1)(rx^r-1)+0.75x^r=0

The Attempt at a Solution

What to do next?

 

[STEMucator]

Homework Statement

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.

Homework Equations

y=x^r
y'=rx^r-1
y"=r(r-1)x^r-2
(x+1)²(r(r-1)x^r-2)+3(x+1)(rx^r-1)+0.75x^r=0

The Attempt at a Solution

What to do next?

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

 

[Success]

No. Please help me from the work that I've shown you.

 

[ehild]

Homework Statement

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.

Homework Equations

y=x^r
y'=rx^r-1
y"=r(r-1)x^r-2
(x+1)²(r(r-1)x^r-2)+3(x+1)(rx^r-1)+0.75x^r=0

The Attempt at a Solution

What to do next?

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

ehild

 

[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 = e^t. 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!

 

[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