- #1
hoch449
- 13
- 0
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]
[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]
Last edited: