Solving second order linear homogeneous differential equation

1. Apr 29, 2012

the0

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

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

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

2. Relevant equations

3. 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!

2. Apr 29, 2012

dikmikkel

Looks like Heun's Differential Equation to me.

3. Apr 29, 2012

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.

4. Apr 29, 2012

HallsofIvy

Staff Emeritus
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
$$\frac{dy}{dx}= \frac{dy}{d\theta}\frac{d\theta}{dx}= \frac{1}{\sqrt{1- x^2}}\frac{dy}{d\theta}$$

5. Apr 29, 2012

HallsofIvy

Staff Emeritus
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.

Last edited: Apr 29, 2012
6. Apr 29, 2012

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?

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