How can i solve the following ODE?
$$y'' y^n + y^{n+1}=1$$
Where $y:=y(x)$ is a differentiable function of $x$ and $y^n := yy ...y$.
Thanks for any help.
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$\begingroup$ One obvious solution is $y=1$ but since it's non-linear I don't see how that helps. $\endgroup$– John Wayland BalesCommented Dec 6, 2016 at 7:30
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$\begingroup$ @Moo I suspect that if $y^{(n)}$ were intended then the first term would have been written $y^{(n+2)}$ $\endgroup$– John Wayland BalesCommented Dec 6, 2016 at 7:33
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$\begingroup$ for $n=2$, wolframalpha gives an answer that's half a page long. seems like a hard problem. $\endgroup$– dezdichadoCommented Dec 6, 2016 at 8:05
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2 Answers
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Transform
$$y''+y=y^{-n},$$
$$y'y''+yy'=y^{-n}y',$$ (provided $y'\ne0$), then integrate $$y'^2+y^2=\frac2{1-n}y^{1-n}+C,$$
and a second time
$$\int\frac{dy}{\sqrt{\frac2{1-n}y^{1-n}+C-y^2}}=x+C'.$$
When $C=3$, there is a nearly human-looking antiderivative https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt(3-y%5E2-2%2Fy)dy
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$\begingroup$ Note, that this neglects the special solutions $y=1$ and $y=-1$ for odd $n$. See comment of @John-Wayland-Bales. $\endgroup$– TobiasCommented Dec 6, 2016 at 8:26
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1$\begingroup$ @Tobias: I focused on the "hard stuff". Special cases such as $n=-1, y'=0$ and the handling of the sign of the square root are not handled. $\endgroup$– user65203Commented Dec 6, 2016 at 8:29
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1$\begingroup$ Then you should at least mention this in the answer. $\endgroup$– TobiasCommented Dec 6, 2016 at 8:30
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The differential equation is reduced to a variable sepearble problem as follows on dividing by $y^n$ if $y\neq0$ :$y''+y=y^{-n}\implies\frac{y''}{y^{-n}-y}=1,y^{-n}\neq y$, which is difficult to be integrated further. However, as Yves Daoust has pointed out it can be integrated once by multiplying the original equation by $y'$. Note, however, that $y=1$ trivially satisfies this equation and $y=-1$ satisfies it for odd $n$, which are the singular solutions to this equation.
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$\begingroup$ There is no resolution method for a separable equation of the second order as there is for the first order. $\endgroup$– user65203Commented Dec 6, 2016 at 8:23
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$\begingroup$ @YvesDaoust so, should I delete the answer? $\endgroup$ Commented Dec 6, 2016 at 8:24
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1$\begingroup$ In this particular case, as there is no instance of $y'$, the equation is in a way "doubly separable" and can be integrated if you multiply by $y'$. So your answer is a little "short". $\endgroup$– user65203Commented Dec 6, 2016 at 8:28