Solve the differential (just understand 1st step)

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The discussion focuses on solving the differential equation \(\frac{d^2 S}{d\theta^2} + \cot\theta\frac{dS}{d\theta} - \frac{m^2}{\sin^2\theta}S(\theta) + \frac{cS(\theta)}{\hbar}=0\). The first step involves substituting \(w = \cos\theta\) to simplify the equation, transforming it into a Confluent Hypergeometric Differential Equation (CHDE). This substitution is essential as it aligns the equation with the standard form of CHDE, facilitating easier solutions. The discussion emphasizes the necessity of adjusting \(S\), \(\frac{dS}{d\theta}\), and \(\frac{d^2 S}{d\theta^2}\) accordingly to eliminate \(\theta\).

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Solve the following differential:

[tex]\frac{d^2}{d\theta^2} + cot\theta\frac{dS}{d\theta} - \frac{m^2}{sin^2\theta}S(\theta) + \frac{cS(\theta)}{\hbar}=0[/tex]

The first step is:

"For convenience we change the independent variable, by making the substitution [tex]w=cos\theta[/tex]"

So my question is:

How do you know that you need to make this substitution? How does this make things easier?

**Also this differential ends up being transformed into a Confluent Hypergeometric Differential Equation** If that helps..

My guess as to why, is that they are trying to set up this differential so it looks like the standard form CHDE which we know how to solve. What do you think?

where standard form is:

[tex]x\frac{d^2u}{dx^2} + (b-x)\frac{du}{dx} + au=0[/tex]
 
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You are missing S in your first term, so your differential equation probably should be :
[tex]\frac{d^2 S}{d\theta^2} + cot\theta\frac{dS}{d\theta} - \frac{m^2}{sin^2\theta}S(\theta) + \frac{cS(\theta)}{\hbar}=0[/tex]

As far as the substitution is concerned, some clever person figured out that it would be a good one to make. If w = cos(theta), then theta = arccos(w), and dw = -sin(theta)d theta. When you make your substitution, it seems to me that you'll need to do something with S = S(theta), dS/d theta, and d^2 S/(d theta)^2 as well, to get rid of theta everywhere it appears.
 

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