Solving second order linear homogeneous differential equation

[the0]

Homework Statement

Find the set of functions from (-1,1)→ℝ which are solutions of:

(x^{2}-1)y''+xy'-4y = 0

Homework Equations

The Attempt at a Solution

OK, I'm not really sure how to go about solving this equation, I have only previously attempted problems where the functions in x are constant.

There is a hint which says to use the change of variable:
x=cos(θ)

doing this I get:

(1): (cos^{2}(θ)-1)y''+cos(θ)y'-4y = 0

which can be rearranged to give:

(2): sin^{2}(θ)y''-cos(θ)y'+4y = 0
or
(3): (\frac{cos(2θ)-1}{2})y''+cos(θ)y'-4y = 0

No idea what to do next!
Any pointers would be great, thank you!

 

[dikmikkel]

Looks like Heun's Differential Equation to me.

 

[the0]

I've never come across that before :/

How can I use that to solve it?

rearrange to give:

y''+\frac{x}{(x+1)(x-1)}y'-\frac{4}{(x+1)(x-1)} = 0 ?

Then I can't see why the change of variable hint has been given.

 

[HallsofIvy]

Homework Statement

Find the set of functions from (-1,1)→ℝ which are solutions of:

(x^{2}-1)y''+xy'-4y = 0

Homework Equations

The Attempt at a Solution

OK, I'm not really sure how to go about solving this equation, I have only previously attempted problems where the functions in x are constant.

There is a hint which says to use the change of variable:
x=cos(θ)

doing this I get:

(1): (cos^{2}(θ)-1)y''+cos(θ)y'-4y = 0

This is completely wrong. You have replaced x with cos(\theta) but you have left y' and y'' as derivatives with respect to x.
If x= cos(\theta) then \theta= cos^{-1}(x) so that
\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}
and
\frac{dy}{dx}= \frac{dy}{d\theta}\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}\frac{dy}{d\theta}

which can be rearranged to give:

(2): sin^{2}(θ)y''-cos(θ)y'+4y = 0
or
(3): (\frac{cos(2θ)-1}{2})y''+cos(θ)y'-4y = 0

No idea what to do next!
Any pointers would be great, thank you!

 

[HallsofIvy]

I've never come across that before :/

How can I use that to solve it?

rearrange to give:

y''+\frac{x}{(x+1)(x-1)}y'-\frac{4}{(x+1)(x-1)} = 0 ?

Then I can't see why the change of variable hint has been given.

This "rearrangement" is wrong.

The correct arrangement is
y''+\frac{x}{(x+1)(x-1)}y'-\frac{4y}{(x+1)(x-1)} = 0

Now, how does that help?

The substitution, x= cos(\theta), if done correctly, simplifies this a great deal.

 

[the0]

OK, thanks a lot!

Right, am I correct in substituting the following:

x = cos(θ)

\frac{dy}{dx} = (\frac{-1}{sin(θ)})\frac{dy}{dθ}

\frac{d^{2}y}{dx^{2}} = (\frac{-cos(θ)}{sin^{3}(θ)})\frac{d^{2}y}{dθ^{2}}

?


If so, I get:

cos(θ)\frac{d^{2}y}{dθ^{2}}-cos(θ)\frac{dy}{dθ}-4sin(θ)y = 0

What next?
Or have I made some stupid error?