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Is there any familiar formula or method to solve an ODE (for $y(x), \ x\in\mathbb{R}$) of the form

$$ (\mu(x)\cdot y'(x))'-\mu(x)\cdot y(x)+r(x)=0 $$ where $\mu(x)$ is real smooth non vanishing function and r(x) is a smooth real function.

Thank you

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First you probably will want to reduce the formula to:

$$\mu y'' + mu' y' -\mu y + r=0$$ $$=y'' + \frac{\mu'}{mu} y - y + \frac{r}{\mu}$$ $$=y'' + (\ln \mu)' y - y + \frac{r}{\mu}=0.$$

Assuming that the homogeneous solution, ie the solution to:

$$y'' + (\ln \mu)' y -y=0$$

is known, then you can just use variation of parameters, since this is a second order linear ODE.

In order to find the solution to the homogeneous equation, we will want to use the change of variables:

$$\ln y = \ln z -\frac{1}{2} \int (\ln \mu)' \mathrm{d} x$$

in order to get the equation:

$$z'' + (-1 -\frac{1}{2}(\ln \mu)'' - \frac{1}{4}[(\ln \mu)']^2)z=0.$$

However, whether this can be solved explicitly will obviously depend on what $(\ln \mu)'$ is.

See: http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html

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