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.
 A: The general form of equation
$$
A^2-B^2=1
$$
can be parametrized as $A=\pm\cosh(u)$, $B=\sinh(u)$ similar to a circle equation. If $(A,B)$ change smoothly, but remain on this curve, then also $u$ is a smooth function.
Here that gives $$y(x)=\pm2\cosh(u(x)), ~~ y'(x)=\sinh(u(x))$$ from that parametrization. Now take the derivative of the first equation and compare with the second, giving $$u'(x)=\pm\tfrac12\implies u(x)=\pm\tfrac12x+c.$$ This already solves the problem completely
$$
y(x)=\pm2\cosh(\tfrac12x+C).
$$

As to your attempt, that is not possible as the equation is not linear. You can get a linear equation by taking the derivative
$$
2y'(4y''-y)=0.
$$
Excluding the constant solutions and solutions with constant segments, the second factor indeed has the general solution
$$
y=c_1e^{\frac12 x}+c_2e^{-\frac12 x}.
$$
Inserting back into the original equation results in
$$
(c_1^2e^{x}-2c_1c_2+c_2^2e^{-x})-(c_1^2e^{x}+2c_1c_2+c_2^2e^{-x})+4=0
\\
\implies c_1c_2=1,~~c_1=\pm e^C,~c_2=\pm e^{-C}
$$
which again produces the solution.

The coefficients in the video give just another parametrization of the coefficient pair. $c_1=\frac12c$ and $c_2=2c^{-1}$ still satisfy $c_1c_2=1$.
A: 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.
