Solving differential equation $af'(x)^2-f''(x)=0$ I want to solve the differential equation $$af'(x)^2-f''(x)=0$$ where $a $ is constant and we have the boundary conditions $f(0)=0, f (1)=c $ for some positive $c$.
If $a=0$ we get $f (x)=cx$ but what if $a\ne 0$. Could someone show how to solve this differential equation?

Set $y=f(x)$. The problem can be rewritten as $a(y')^2-y''=0$. Set $u:=y'$, so we get $au^2-u'=0\Rightarrow u'=au^2\Rightarrow \int \frac {du}{u^2}=\int adx\Rightarrow -\frac 1u=ax+C\Rightarrow u=-\frac {1}{ax+C}$, or by setting $c:=-C$ we get $u=\frac {1}{c-ax}$. Finally we have $y'=\frac {1}{c-ax}\Rightarrow\int dy=\int \frac {dx}{c-ax}\Rightarrow y=-\frac 1a ln|c-ax|+K$, where $K$ a real constant.
