Non-linear differential equation with variable coefficient Given $$ x\dfrac{d^2x}{dt^2}+\left (\dfrac{dx}{dt}\right )^2=x\dfrac{dx}{dt},\quad x(0)=0, x(1)=1.$$ What is the value of $x(2)$? I tried with $dx/dt=y, dy/dt=(dy/dx)(dx/dt)$ but failed at integration.
 A: Even if this is not a direct answer but it leads to the answer.
This equation can be written as 
$$ \dfrac{d}{dt} \left( x\dfrac{dx}{dt} \right)=x\dfrac{dx}{dt},\quad $$
Therfore     $ \quad x\dfrac{dx}{dt}=ce^t   $ 
Then
$$ \int x dx = \int ce^t dt$$
so $$\frac{1}{2}x^2(t)=ce^t+d$$
Can you take it from here?
A: You can divide the equation by $x\dot{x}$ and integrate the equation (chain rule).
$$  \frac{\ddot{x}}{\dot{x}} +  \frac{\dot{x}}{x} = 1 $$
$$ \implies \ln(\dot{x}) + \ln(x) =  \ln(x\dot{x})  = t+ c  $$
$$ \implies x\dot{x} = ke^{t}  $$
The last equation is separable again.
A: Let$p=\frac{dx}{dt}$ then $\frac{d^2x}{dt^2}=p \frac{dp}{dx}$ and the equation becomes 
\begin{eqnarray*}
\frac{dp}{dx}=\frac{x-p}{x}
\end{eqnarray*}
or $p=0$. The equation above is easily solved by the substitution $p=xu$ ... good luck.
A: $$x\dfrac{d^2x}{dt^2}+\left (\dfrac{dx}{dt}\right )^2=x\dfrac{dx}{dt},\quad x(0)=0, x(1)=1.$$
$$xx''+(x')^2=x'x$$
$$\implies (x'x)'=x'x$$
Substitute $z=x'x$
$$z'= z$$
This last equation is separable...
