Why using $p=y'$ doesn't lead to answer in $y^2(1+y'^2)=4$? I have the ODE $y^2(1+y'^2)=4$ to solve this I used the substitution $y'=p$
$$y^2(1+p^2)=4$$
$$2y(1+p^2)dy+2py^2dp=0$$
$$(p^2+1)dy+py\;dp=0$$
$$\frac{dy}y+\frac{p}{p^2+1}dp=0$$
$$\ln|y|+\frac12\ln|p^2+1|=\ln|c|$$
$$y\sqrt{p^2+1}=c$$
Using $p^2+1=\frac4{y^2}$, I get $2=c$ ! I can't find my mistake.
 A: You didn't make a mistake, but you have arrived back to the original equation. Note that $c$ can also be $-2$.
$$y\sqrt{p^2+1}=\pm2\implies y^2(1+p^2)=4$$
Instead, note that the differential equation is separable. It can be written as
$$y'=\frac{\sqrt{4-y^2}}{y}\quad\text{or}\quad y'=-\frac{\sqrt{4-y^2}}{y}$$
A: I think you've only circulated in the differential equation. You have written the  series of steps only to arrive at the restatement of the original differential equation.
From your solution,
$$y\sqrt{p^2-1} = c$$
or $$ y^2 (p^2-1) = c^2$$
which is restatement of the original equation and $c^2 = 4$ only satisfies the condition of the original differential equation. Therefore you need to try to solve it in such a way that 4 does not get removed. One way could be to rewrite it as follows:
$$ y'^2= \frac{4}{y^2} - 1$$
This will lead to solution.
A: Alternative solution strategy: Expanding the left side, you get a circle equation
$$
y^2+(yy')^2=4
$$
that can be parametrized as $y=2\sin u$, $yy'=2\cos u$ which then leads to
$$
y'=2u'\cos u\implies 2\cos u=yy'=4u'\sin u\cos u\implies x+c=-2\cos u
$$
if $\cos u\ne 0$. Then
$$
y^2=4(1-\cos^2u)=4-(x+c)^2.
$$
