ODE Solution $y''(x) = y^4(x)$

Question

Is it possible to analytically calculate the ODE $y''(x) = y^4(x)$? It's easy to see that $y \equiv 0$, but is there another non-trivial solution to this problem?

If not, is there any software to help me with this problem?

Answer 1

You can immediately integrate once by multiplying by $y':$ \begin{align*} y''&=y^4\\ y''y'&=y^4y'\\ \frac{(y')^2}{2}&=\frac{y^5}{5}+C_1\\ y'&=\pm\sqrt{\frac{2y^5}{5}+2C_1}\\ \int\frac{dy}{\sqrt{\dfrac{2y^5}{5}+2C_1}}&=\pm(x+C_2). \end{align*} Now the integral on the LHS you can write in terms of hypergeometric functions as follows: $$\frac{y\sqrt{1+\dfrac{2y^5}{5C_1}}\;\;_2F_1\!\!\left(\dfrac{1}{5},\dfrac{1}{2};\dfrac{6}{5};-\dfrac{2y^5}{5C_1}\right)}{\sqrt{\dfrac{2y^5}{5}+C_1}}=\pm(x+C_2).$$ This is basically solved for $x.$ I highly doubt it's possible to solve for $y.$