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?
 A: $$af' = \frac{f''}{f'}$$
$$af' = \ln f'$$
$ay+b\log(y)+c=0$
$\implies y = W$ where W is Lambert's W function
i.e
$$\begin{align*}
f' 
& = W\\
&f = \int W
\end{align*}$$
A: \begin{gather*}
Let\ f'( x) =u( x)\\
Then\ f''( x) =u'( x)\\
( We\ will\ refer\ to\ u( x) \ as\ simply\ u)\\
The\ equation\ becomes\\
a\cdotp u^{2} =\frac{du}{dx}\\
a\cdotp dx=\frac{du}{u^{2}}\\
Integrate\ both\ sides\ to\ solve\ the\ equation.\\
Let\ the\ solution\ obtained\ from\ solving\ this\ equation\ be\ w( x)\\
Then,\ f( x) =\int w( x) dx
\end{gather*}
Did this help?
A: 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.
A: The differential equation can be written as
$$
\frac{f''(x)}{f'(x)^2} = a.
$$
Taking antiderivatives gives
$$
-\frac{1}{f'(x)} = ax + B,
$$
for some constant $B.$ Thus,
$$
f'(x) = -\frac{1}{ax+B}
$$
and taking another antiderivative gives
$$
f(x) = C - \frac{1}{a}\ln(ax+B)
$$
where $C$ is a constant.
Boundary condition $f(0)=0$ then gives
$$
0 = f(0) = C - \frac{1}{a}\ln(a\cdot 0+B) = C - \frac{1}{a}\ln B,
$$
i.e.
$$
C = \frac{1}{a}\ln B.
$$
Boundary condition $f(1)=c$ gives
$$
c = f(1) = C - \frac{1}{a}\ln(a\cdot 1+B) = \frac{1}{a}\ln B - \frac{1}{a}\ln (a+B) = \frac{1}{a}\ln\frac{B}{a+B},
$$
from which we can solve for $B$ and get
$$
B = \frac{a}{e^{-ac} - 1}.
$$
Thus we get the solution
$$
f(x) = \frac{1}{a}\ln \frac{a}{e^{-ac} - 1} - \frac{1}{a}\ln(ax+\frac{a}{e^{-ac} - 1}) \\
%= \frac{1}{a} \ln \frac{\frac{a}{e^{-ac} - 1}}{ax+\frac{a}{e^{-ac} - 1}}
= \frac{1}{a} \ln \frac{1}{x(e^{-ac} - 1)+1}.
$$
