# Differential equation: $A(x)y''(x)+A'(x)y'(x)+y(x)/A(x)=0$

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 ...

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 $$\int \frac{1}{\sqrt{\lambda-y^2}} dy = \pm\int \frac{1}{A}dx$$
• 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. – Chinny84 Dec 20 '14 at 9:39