Non-linear differential equation with boundary conditions

Question

Given $\dfrac{d^2y}{dx^2}+\dfrac{1}{y}\left (\dfrac{dy}{dx}\right)^2-\dfrac{dy}{dx}=0,\quad y(0)=0, y(1)=1.$ What is $y(2)$? A rearrangement gives $\left (\dfrac{1}{y'}\right )'+\dfrac{1}{y'}=y$. I do not know how to proceed further.

Answer 1

From $yy''+(y')^2-yy'=0$ and $z:=y^2$ you get

$$z''-z'=0.$$

Answer 2

You made a little mistake. It should be $$\left (\dfrac{1}{y'}\right )'+\dfrac{1}{y'}=\frac 1y$$

Another approach $$\dfrac{d^2y}{dx^2}+\dfrac{1}{y}\left (\dfrac{dy}{dx}\right)^2-\dfrac{dy}{dx}=0,\quad y(0)=0, y(1)=1.$$ $$y''+\dfrac{1}{y}(y')^2=y'$$ Multiply by $y (y \neq 0)$ $$yy''+(y')^2=y'y$$ This equation is separable $$(y'y)'=y'y$$ Simply integrate $$\int \frac { d(y'y)}{y'y}=\int dx$$ $$\ln(y'y)=x+K \implies y'y=Ke^x$$ $$\frac 12(y^2)'=Ke^x \implies (y^2)'=Ce^x$$ $$y(x)=\pm\sqrt {c_1e^x+c_2}$$ $$\boxed {y(x)=k_1\sqrt {e^x+k_2}}$$ Apply the initial condition to find $k_1,k_2$