Hard Differential Equation Can anyone help me to find the solution of this ODE : $$4(y')^2-y^2+4=0.$$
I've tried to find it's solution by putting $y = e^{at}$ (for null solution) and $y = 2$ (for particular solution). My final solution is $$y = c_1 e^{0.5t} + c_2 e^{-0.5t} + 2,$$
but didn't match with the solution of this dude in this video link [Time stand 2:20 ] "when question is shown in the video".
Even I've plotted his solution and my solution in graph plotter but there is slight difference in the graphs.
Please, explain in full detail if you know how.

Edit: To look for a constant solution, $y' = 0, \implies y = \pm2$
For a general solution, you can rewrite this as:
$$4(y')^2 = y^2-4 \implies y' = \left(\sqrt{y^2-4}\right)/2$$
$$\implies \int\frac{\,dy}{\sqrt{y^2-4}} = \int\,dx/2$$
Put: $y \rightarrow 2\sec z \implies dy = 2\sec z\tan z \,dz$
$$\implies \int \sec z \,dz = x/2 + C$$
$$\implies \ln|\sec z + \tan z| = x/2 + C$$
Now plug $\sec z = y/2 $ back in this.
