Non-linear differential equation with variable coefficient

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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.

Answer 4

$$x\dfrac{d^2x}{dt^2}+\left (\dfrac{dx}{dt}\right )^2=x\dfrac{dx}{dt},\quad x(0)=0, x(1)=1.$$ $$xx''+(x')^2=x'x$$ $$\implies (x'x)'=x'x$$ Substitute $z=x'x$ $$z'= z$$ This last equation is separable...