Differential equation for non-ordinary homogenous equation

Question

What is the solution to the differential equation: $$y''+y^\alpha=0$$ where $\alpha\neq 1,0$?

Tried finding solutions on websites with differential equation calculators but they gave nothing. So does it even exist?

Answer 1

$$\frac{d^2y}{dt^2} + y^a = 0 \Leftrightarrow \frac{d^2y}{dt^2}=-y^a \Rightarrow \frac{dy}{dt}\frac{d^2y}{dt^2}=-\frac{dy}{dt}y^a \Rightarrow \int\frac{dy}{dt}\frac{d^2y}{dt^2}dt=\int-\frac{dy}{dt}y^adt $$

$$\Rightarrow$$

$$\frac{1}{2}\bigg(\frac{dy}{dt}\bigg)^2 = c_1 - \frac{y^{a+1}}{a+1} \Rightarrow \bigg(\frac{dy}{dt}\bigg)^2 = 2c_1 - \frac{2y^{a+1}}{a+1}$$

$$\Rightarrow$$

$$\frac{dy}{dt} = \pm \sqrt{2c_1 - \frac{2y^{a+1}}{a+1}} \Leftrightarrow \frac{\frac{dy}{dt}}{\sqrt{2c_1 - \frac{2y^{a+1}}{a+1}}} = \pm 1$$

$$\Rightarrow$$

$$\int\frac{\frac{dy}{dt}}{\sqrt{2c_1 - \frac{2y^{a+1}}{a+1}}}dt = \pm \int dt$$

This is as far as you can get to approaching a solution without the use of special functions.

Extending it a bit :

$$\Rightarrow \frac{_2F_1\big(\frac{1}{2}, \frac{1}{a+1};\frac{1}{a+1};\frac{y^{a+1}}{c_1 + ac_1}\big)\sqrt{\frac{c_1+ac_1-y^{a+1}}{c_1 + ac_1}}y(t)}{\sqrt{2c_1 - \frac{2y^{a+1}}{a+1}}}= c_2 \pm t$$

where $_2F_1(\dots)$ is the hypergeometric function.