Solving this ODE (Euler's form?)

Question

I am having a struggle solving this ODE:

$$(1-y)y''+2y'^2=0$$

I tried to solve this by changing it to Euler's form, but could not go any further. Can anyone solve this for me? And if possible, can you show me how to solve the equation by step? Thank you.

Answer 1

Following Moo's hint,

Here $' \equiv \frac{d}{dx}$. Let $v = y'$. Then $y'' = v' = \frac{dv}{dy} \cdot \frac{dy}{dx} = \frac{dv}{dy} v$, \begin{align} (1 - y)y'' + 2(y')^2 &= 0\\ \Rightarrow (1 - y)\frac{dv}{dy}v + 2v^2 &= 0\\ \Rightarrow v\left((1 - y)\frac{dv}{dy} + 2v \right) &= 0 \end{align} Hence, we have $v = 0$ or $\left((1 - y)\frac{dv}{dy} + 2v \right) = 0$.

If $v = 0$, then $y(x) = c_1$ for some $c_1 \in \mathbb{R}$.

If $\left((1 - y)\frac{dv}{dy} + 2v \right) = 0$, then \begin{align} (y - 1)\frac{dv}{dy} &= 2v\\ \Rightarrow \int \frac{1}{2v} dv &= \int \frac{1}{y - 1} dy\\ \Rightarrow v &= c_2(y - 1)^2,\ \exists c_2 \in \mathbb{R}\\ \Rightarrow y' &= c_2(y - 1)^2 \\ \Rightarrow \int \frac{1}{(y - 1)^2} dy &= c_2 \int dx\\ \Rightarrow \frac{-1}{y - 1} &= c_2 x + c_3,\ \exists c_3 \in \mathbb{R}\\ \Rightarrow y &= \frac{1}{c_2 x + c_3} + 1\\ \end{align}