Non-Linear ordinary differential equation involving first derivative

Question

Kindly help me in solving non-linear ODE: $$y'^2+y^2+4=0$$

Answer 1

Substitute $y=2\sinh u$ therefore $$4u'^2\cosh^2u+4+4\sinh^2u=0$$which means that $$4u'^2\cosh^2u+4\cosh^2u=0$$since $\cosh(.)$ is always positive we obtain $$u'^2+1=0$$which has no real answer, but if whole the complex plane is included then$$u'=\begin{cases}i& ,x\in S\\-i& ,x\notin S\end{cases}$$where $i=\sqrt {-1}$ and at least one of $S$ or $S^c$ is homeomorphic to $\Bbb N$ (you don't need to mind this constraint so much, just choose $S$ so that $u$ is integrable). One answer is $$u=\begin{cases}ix& ,x\in S\\-ix& ,x\notin S\end{cases}$$which leads to $$y=\begin{cases}i\sin x& ,x\in S\\-i\sin x& ,x\notin S\end{cases}$$