# $y''=a\frac{y}{(y')^2+b},y(0)=0,y'(0)=c,\forall c\in\mathbb{R}$ [closed]

Can some solve the following equation or at least tell me how to approximate the result? $$y''=a\frac{y}{(y')^2+b},y(0)=0,y'(0)=c,\forall c\in\mathbb{R}$$ I have no idea if it is an easy task, but for my level of math, I can't solve it and I need it for a project. Wolfram cannot give me an answer, at least without pro computation power. It tells me it is a second order non linear ODE. Thanks in advance

• If Wolfram couldn't solve it, we probably can't. Nov 29, 2023 at 23:21

The problem can be, at least, reduced to quadrature. Let $$p=y'$$; then the ODE can be rewritten as \begin{align} y''=\frac{dp}{dx}=p\frac{dp}{dy}=\frac{ay}{p^2+b} &\implies \int (p^2+b)p\,dp=\int ay\,dy \\ & \implies \frac{p^4}{4}+\frac{bp^2}{2} = \frac{ay^2}{2}+C_1. \tag{1} \end{align} From the initial condition $$y(0)=0, p(0)=y'(0)=c$$ it follows that $$C_1=\frac{c^4}{4}+\frac{bc^2}{2}$$.
Solving $$(1)$$ for $$p$$ is straightforward, and we find $$p=\pm\sqrt{-b\pm\sqrt{(b+c^2)^2+2ay^2}}. \tag{2}$$ The signs in $$(2)$$ must be chosen in such a way that $$p(0)=c$$ is satisfied. Thus, we have four cases to consider: $$p(y)=\begin{cases} +\sqrt{-b+\sqrt{(b+c^2)^2+2ay^2}}&\text{if b+c^2\geq0 and c\geq0,} \\ +\sqrt{-b-\sqrt{(b+c^2)^2+2ay^2}}&\text{if b+c^2<0 and c\geq0,} \\ -\sqrt{-b+\sqrt{(b+c^2)^2+2ay^2}}&\text{if b+c^2\geq0 and c<0,} \\ -\sqrt{-b-\sqrt{(b+c^2)^2+2ay^2}}&\text{if b+c^2<0 and c<0.} \end{cases} \tag{3}$$ Integrating $$y'=p(y)$$ we finally obtain $$x=\int_0^y\frac{du}{p(u)}, \tag{4}$$ which yields the solution to the ODE in implicit form.