How do you solve for f in the differential equation (x^2-a^2)f'+xf=0?

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SUMMARY

The differential equation (x² - a²)f' + xf = 0 can be solved using separation of variables. By rearranging the equation, we obtain the form f' / f = -x / (x² - a²). Integrating this leads to the solution ln(f) = -0.5 ln(x² - a²) + C, where C is the constant of integration. This method effectively simplifies the problem and provides a clear path to the solution.

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Helios
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This one looks easy but I don't know.

( x^{2} - a^{2} )f' + xf = 0
 
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Isn't that separable?
 
Rearrange to get

\frac{f^{\prime}}{f} = - \frac{x}{x^2 - a^2}

now integrate to get

\ln{f} = -\frac{1}{2} \ln{(x^2 - a^2)} + C
 

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