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So give the differential equation

$$A(x)y''(x)+A'(x)y'(x)+\frac{y(x)}{A(x)}=0,$$ with $A(x)$ a known function and $y(x)$ te be determined. What is the solution for this differential equation ?

I've tried substituting $y(x)=A(x)u(x)$, but unfortunately this didn't eliminate my unknown variable $A(x)$. I don't know if there are any other tricks or substitutions that I can try to solve this situation ?

I als considered switching to $u=A(x)$ as my independent variable, but that also didn't help me that much ...

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Rearrange the equation to yield $$ A^2y'' + AA'y' + y = 0 = A^2y'' + \left(\frac{A^2}{2}\right)'y' + y $$ mutiply by $y'$ we find $$ A^2y''y' + \left(\frac{A^2}{2}\right)'y'^2 + yy' = 0\\ \frac{A^2}{2}\left(y'^2\right)' + \left(\frac{A^2}{2}\right)'y'^2 + yy' = 0 $$ the last equation can be written as $$ \frac{1}{2}\dfrac{d}{dx}\left(A^2y'^2\right) + \frac{1}{2}\left(y^2\right)' = 0 $$ or $$ \left(A^2y'^2\right) + y^2 = \lambda $$ so $$ y' = \frac{\pm\sqrt{\lambda-y^2}}{A} $$ or $$ \int \frac{1}{\sqrt{\lambda-y^2}} dy = \pm\int \frac{1}{A}dx $$

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  • $\begingroup$ Very nice solution, I will check this one first thing in the moring! Do you have any tips on to see how this substitution becomes obvious ? $\endgroup$ – Nick Dec 19 '14 at 22:25
  • $\begingroup$ It only became obvious after I multiplyby the denominator of y. Since I saw that I could potential use the product rule but I was missing a half factor and even then I would have an integral for y with respect to x which is unhelpful, so mutiple by $y'$ did the trick to reduce the order. $\endgroup$ – Chinny84 Dec 20 '14 at 9:39

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