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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

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

1 Answer 1

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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.

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