# hint for solving an ordinary differential equation

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.

• One obvious solution is $y=1$ but since it's non-linear I don't see how that helps. Commented Dec 6, 2016 at 7:30
• @Moo I suspect that if $y^{(n)}$ were intended then the first term would have been written $y^{(n+2)}$ Commented Dec 6, 2016 at 7:33
• for $n=2$, wolframalpha gives an answer that's half a page long. seems like a hard problem. Commented Dec 6, 2016 at 8:05

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

• Note, that this neglects the special solutions $y=1$ and $y=-1$ for odd $n$. See comment of @John-Wayland-Bales. Commented Dec 6, 2016 at 8:26
• @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.
– user65203
Commented Dec 6, 2016 at 8:29
• Then you should at least mention this in the answer. Commented Dec 6, 2016 at 8:30

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.

• There is no resolution method for a separable equation of the second order as there is for the first order.
– user65203
Commented Dec 6, 2016 at 8:23
• @YvesDaoust so, should I delete the answer? Commented Dec 6, 2016 at 8:24
• 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".
– user65203
Commented Dec 6, 2016 at 8:28