# Reduction of order of non-linear 2nd order ODE

I am trying to follow the derivation on this wolfram page in which the parametric equations for the brachistochrone problem with friction is found. I am having trouble being able to move from step (29) to (30), in which this 2nd order non linear ODE:

$$[1+y'^2 ](1+\mu y')+2(y-\mu x)y''=0\tag{29}$$

is reduced to this form:

$$\frac{1+(y')^2}{(1+\mu y')^2} = \frac{C}{y-\mu x}\tag{30}$$

I am asking for help in understanding how this reduction is carried out.

## 1 Answer

The derivative of $$y-\mu x$$ is $$y'-μ$$. To transform the equation into a form that has $$F(y')y''$$ on one side and something integrable on the other side, this suggests to try $$-\frac{y'-μ}{y−μx}=\frac{2(y'-μ)y''}{[1+y'^2](1+μy')}$$ Now perform partial fraction decomposition for the right side, $$\frac{2(y'-μ)}{[1+y'^2](1+μy')}=\frac{A+By'}{1+y'^2}+\frac{C}{1+μy'} \\\iff\\ 2(y'-μ)=(A+By')(1+μy')+C[1+y'^2]$$ which implies $$C=-μB$$, $$2-B=μA$$ and $$-(2-B)μ=A$$ giving $$A=0$$, $$B=2$$ and $$C=−2μ$$. Thus after integration of the first equation $$-\ln|y−μx|=\ln|1+y'^2|-2\ln|1+μy'|+c.$$