# Homework Help: Change of Variable With Legendre Equation

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1. Oct 9, 2015

### Bassa

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

Change the independent variable from x to θ by x=cosθ and show that the Legendre equation

(1-x^2)(d^2y/dx^2)-2x(dy/dx)+2y=0

becomes

(d^2/dθ^2)+cotθ(dy/dθ)+2y=0

2. Relevant equations

3. The attempt at a solution

I did get the exact form of what the equation should become, but I had a 2 in front of the cotθ term. I was wondering if the answer that the book provides is missing a factor of 2 in front of the cotθ term?

2. Oct 9, 2015

### MisterX

No, there shouldn't be a "2" factor there.

3. Oct 10, 2015

### Bassa

Thank you very much for your reply! I have spend many an hour trying to figure out where the 2 goes, but that was to to avail. I have the following:

(dy/dx)=(dy/dθ)(-1/sqrt (1-x^2))=(dy/dθ)(-1)(sin(θ))^(-1)

Can you please direct me to where I went wrong on the above statement?

Thanks!

4. Oct 10, 2015

### vela

Staff Emeritus
The problem lies elsewhere. Show your calculation of $(1-x^2)y''$.

5. Oct 10, 2015

### Bassa

(d^2y/dx^2) = (d^2y/dθ^2)(1-x^2)^(-1) = (d^2y/dθ^2)(1-(cosθ)^2)^(-1)

(1-x^2)y'' = (1-(cosθ)^2)(d^2y/dθ^2)(1-(cosθ)^2)^(-1) = (d^2y/dθ^2)

6. Oct 10, 2015

### vela

Staff Emeritus
Differentiate the result you got for y'.
$$\frac{d^2}{dx^2}y = \frac{d}{dx} y' = -\csc\theta \frac{d}{d\theta} y' = -\csc\theta \frac{d}{d\theta} \left(-\csc\theta \frac{dy}{d\theta}\right)$$ You need to use the product rule.

7. Oct 10, 2015

### Bassa

I do have that in my solution and it leads to the earlier reply I posted. It also leads to the correct first term of the right answer. Is the following correct?

-csc(θ)(d/dθ)[y] = (cscθ)^2(d^2y/dθ^2)

8. Oct 10, 2015

### Bassa

I do have that in my solution and it leads to the earlier reply I posted. It also leads to the correct first term of the right answer. Is the following correct?

-csc(θ)(d/dθ)[y] = (cscθ)^2(d^2y/dθ^2)

9. Oct 10, 2015

### vela

Staff Emeritus
No, that's not correct. As I noted earlier, you have to use the product rule. The $d/d\theta$ on the left acts on everything to its right.

10. Oct 12, 2015

### Bassa

Thank you very much! Sorry for the late reply, but I had to sleep on it. It worked perfectly fine! I was inventing my own incorrect rules for calculus, and that did not work very well!

P.S. I was wondering how you were able to type those derivatives very clearly. It is much more clear than having slashes and asterisks?

11. Oct 12, 2015

### PWiz

12. Oct 12, 2015

### Bassa

Thanks a lot!