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I have an equation of the the form $$-\Phi=(\frac{1}\gamma)^2\frac{d} {d\gamma}(\gamma^2 d\Phi/d\gamma) $$ to solve for $\Phi$. It can apparently be solved in two ways; one of which being to substitute $\Phi=f(\gamma)/\gamma$ and find the differential for $f(\gamma)$. The result should come out in terms of sin or cos, but I honestly don't know how.

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Can you include in the question what you get when you substitute $\Phi=f(\gamma)/\gamma$? –  Rahul Apr 12 '12 at 17:40
    
either $\Phi=A \frac{\sin\gamma}\gamma$ or $\Phi=B \frac{\cos\gamma}\gamma$ –  Tan Apr 12 '12 at 18:29
    
Then what's the problem? –  Rahul Apr 12 '12 at 18:48
    
Sorry; that's the result I expect to find, I just don't know how to get there. As far as I can tell, I get as far as $f''/\gamma -f'/\gamma^2 - f/\gamma=0$. Then on I'm stuck. –  Tan Apr 12 '12 at 19:19
    
You've made a mistake in your derivation; you shouldn't get the $f'/\gamma^2$ term, and the $f/\gamma$ term shouldn't have a negative sign. If you had included your derivation in the question like I had said in the first place, we could tell you where you went wrong. –  Rahul Apr 12 '12 at 20:17

2 Answers 2

Work directly!

The differential equation is $\gamma\Phi''+2\Phi'+\gamma\Phi=0$. Now, $(\gamma\Phi)''=\gamma\Phi''+2\Phi'$, hence this is equivalent to $(\gamma\Phi)''+\gamma\Phi=0$. The solutions of $u''+u=0$ are $u(\gamma)=A\cos(\gamma)+B\sin(\gamma)$ hence $\Phi(\gamma)=(A\sin(\gamma)+B\cos(\gamma))/\gamma$.

Note that the only solutions defined at $\gamma=0$ are $\Phi(\gamma)=A\sin(\gamma)/\gamma$.

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Hint :

substitute : $\Phi \cdot \gamma^2=u$ , where $u$ is function in terms of $\gamma$ .

So you get second order linear homogenous ODE :

$$u''-\frac{2}{\gamma}u'+\frac{2+\gamma^2}{\gamma^2}u=0$$

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