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.
How do I solve this differential equation?
1
$\begingroup$
calculus
ordinary-differential-equations
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0Can you include in the question what you get when you substitute $\Phi=f(\gamma)/\gamma$? – 2012-04-12
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0either $\Phi=A \frac{\sin\gamma}\gamma$ or $\Phi=B \frac{\cos\gamma}\gamma$ – 2012-04-12
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0Then what's the problem? – 2012-04-12
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0Sorry; 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. – 2012-04-12
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0You'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. – 2012-04-12
2 Answers
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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 :
$$u''-\frac{2}{\gamma}u'+\frac{2+\gamma^2}{\gamma^2}u=0$$