# How to solve $y^{\prime\prime}=y^2-y, y^{\prime}=0$ when $y(x_0)=1, y^{\prime}(x_0)=2$ where $x_0=\sqrt{2}$

I use $v=y^{\prime}$ and get a first order non-linear ODE, then how to solve $(y^{\prime})^2=\frac{2}{3}y^3-2y^2+\frac{4}{3}$?

• Something is wrong here. The condition $y'=0$ means that the function is constant, and makes everything trivial. Also if $y'=0$, then how can $y'(x_0)$ be $2$? – A. Pongrácz Aug 8 '18 at 9:32

$(y^{\prime})^2=\frac{2}{3}y^3-2y^2+\frac{4}{3}$ is false.

The correct way is with $\frac{dy}{dx}=v(y)$ and $\frac{d^2y}{dx^2}=\frac{dv}{dy}\frac{dy}{dx}=v\frac{dv}{dy}=v\,v'$ $$vv'=y^2-y$$ $$v^2=\int(2y^2-2y)dy$$ $$(y')^2=\frac23 y^3-y^2+c$$

OR EQUIVALENTLY : $$y''=y^2-y$$ $$2y''y'=2y^2y'-2yy'$$ $$(y')^2=\frac23 y^3-y^2+c$$

Condition :

$y'(x_0)=2$ and $y(x_0)=1$ implies $2^2=\frac23-1+c$ hence $c=\frac{13}{3}$ $$y'=\sqrt{\frac23 y^3-y^2+\frac{13}{3}}$$ $$\int\frac{dy}{\sqrt{\frac23 y^3-y^2+\frac{13}{3}}}=x+C$$ $C$ has to be determined according to the condition $y(x_0)=1$. $$\int_1^y\frac{d\theta}{\sqrt{\frac23 \theta^3-\theta^2+\frac{13}{3}}}=x-x_0$$

This is an elliptic integral. One cannot go further with elementary functions.

If it is an academic exercise supposing a solution with elementary functions one can suspect that there is a mistake in the wording of the problem.