Change of Variable With Legendre Equation

[Bassa]

Homework Statement

[/B]
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. Homework Equations

The Attempt at a Solution

[/B]
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?

 

[MisterX]

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

 

[Bassa]

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

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!

 

[vela]

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

 

[Bassa]

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

Thanks for your reply! Here is my calculation for (1-x^2)y'':

(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)

 

[vela]

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.

 

[Bassa]

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.

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)

 

[Bassa]

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.

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)

 

[vela]

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.

 

[Bassa]

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.

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?

 

[PWiz]

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?

It's done using Latex. You can learn how to use it here: https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517

 

[Bassa]

Thanks a lot!