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.


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 $$ -\ln|y−μx|=\ln|1+y'^2|-2\ln|1+μy'|+c. $$


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