solution of $y^{\prime \prime} + y^n = 1$

Question

I am not able to figure out the solution for the differential solution $$y^{\prime \prime} + y^n = 1$$ I want to specifically find an answer for $$y^{\prime \prime} + y^2= 1$$and $$y^{\prime \prime} + y^3 = 1$$

If anyone can help!

Answer 1

$$y''+y^n=1$$ An obvious solution is $y=1$. Apart this trivial case : $$2y''y'+2y^ny'=2y'$$ $$(y')^2+\frac{2}{n+1}y^{n+1}=2y+c_1$$ $$y'=\frac{dy}{dx}=\pm\sqrt{2y+c_1-\frac{2}{n+1}y^{n+1}}$$ $$x=\pm\int\frac{dy}{\sqrt{2y+c_1-\frac{2}{n+1}y^{n+1}}}+c_2$$ $x(y)$ is the inverse function of $y(x)$.

There is no general closed form for the integral and even more for the inverse function.

Further calculus supposes to analytically solve a polynomial equation of $(n+1)$ degree. Theoretically it is possible up to $n+1=4$. So, the closed form exists for $n=2$ and $n=3$ not for $n>3$.

In fact, the main difficulty comes from the analytical solving of the polynomial equation, which roots involve huge formulas. If we suppose that the roots are known, further calculs involves elliptic integral of the first kind to express $x(y)$, then Jacobi amplitude function to express y(x) :