# Change of independent variables

1. Jan 2, 2008

### jesuslovesu

[SOLVED] Change of independent variables

1. The problem statement, all variables and given/known data
x = cos$$\theta$$
Show
$$(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0$$
becomes
$$\frac{d^2y}{d\theta ^2}+ cot(\theta) \frac{dy}{d\theta} + 2y = 0$$

2. Relevant equations

3. The attempt at a solution

dx = -sin($$\theta$$) d$$\theta$$
...
As a total stab in the dark I've tried substituting -sin($$\theta$$) d$$\theta$$ in for dx, but that didn't seem to work... and I don't think that is mathematically correct to begin with. How do I get the relationship between dy/dx and dy/dtheta? if y were given I think it would be a lot easier

2. Jan 2, 2008

### CompuChip

Use this (let me call it "inverse chain rule")

$$\frac{dy}{dx} = \frac{dy}{d\theta} \frac{d\theta}{dx}$$
where you can explicitly calculate the latter factor from the equation you wrote under (3).

Once you have that, you can do the same trick again:
$$\frac{d^2y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} = \left( \frac{d}{d\theta} \frac{dy}{dx} \right) \frac{d\theta}{dx}$$

Last edited: Jan 2, 2008
3. Jan 2, 2008

### Defennder

Did you leave out a factor of 2 for the coefficient of cot (theta) in the answer?

Should it be $$\frac{d^2y}{d\theta ^2}+ 2cot(\theta) \frac{dy}{d\theta} + 2y = 0$$ ?

4. Jan 2, 2008

### jesuslovesu

Hey guys thanks for your replies, I'm trying it out right now.
Defennder: Nope, it's just cot

5. Jan 2, 2008

### jesuslovesu

Got it, thanks

Last edited: Jan 2, 2008