How I can find all solutions of the ODE $(y')^{2}+y^{2}=4$ I want to find all solutions of this ordinary differential equation:
$$
 (y')^{2}+y^{2}=4
$$ 
but I don't know how. It is impossible by use of series method or Laplace transform?
 A: Hint: 
$$y' = \pm \sqrt{4-y^2} \Rightarrow \int \frac{\mathrm{d}y}{\sqrt{4-y^2}} = \int \pm \, \mathrm{d}x$$
The LHS is a standard $\arcsin$ integral. 
A: To complement other answers observe that $y=\pm2$ are constant solutions. There are more solutions composed of pieces of $\sin$, $\cos$ and the constant solutions, like
$$
y(x)=\begin{cases}2\sin x & x\le\pi/2,\\2 & x>\pi/2.\end{cases}
$$
The reason for this is that $\sqrt{4-y^2}$ is not Lipschitz at $y=\pm2$ and there is no uniqueness of solution when the initial value is $\pm2$.
A: $$(y\prime )^2+y^2 =4\\ y^\prime =\pm \sqrt { 4-y^2 } \\ \int \frac { dy }{ \sqrt { 4-y^2 }  } =\pm \int dx \\ y=2\sin z \\ \int { \frac { 2\cos z \, dz }{ 2\left| \cos z  \right|  }  } =\pm x+C\\ \\ \arcsin \frac y 2 =x+c $$
A: $$y'(x)^2+y(x)^2=4\Longleftrightarrow$$
$$y'(x)=\pm\sqrt{4-y(x)^2}\Longleftrightarrow$$
$$\frac{y'(x)}{\sqrt{4-y(x)^2}}=\pm1\Longleftrightarrow$$
$$\int\frac{y'(x)}{\sqrt{4-y(x)^2}}\space\text{d}x=\int\pm1\space\text{d}x\Longleftrightarrow$$
$$\int\frac{y'(x)}{\sqrt{4-y(x)^2}}\space\text{d}x=\text{C}\pm x\Longleftrightarrow$$

For the integral on the LHS:
Substitute $u=y(x)$ and $\text{d}u=y'(x)\space\text{d}x$:

$$\int\frac{1}{\sqrt{4-u^2}}\space\text{d}u=\text{C}\pm x\Longleftrightarrow$$
$$\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}\space\text{d}u=\text{C}\pm x\Longleftrightarrow$$

Substitute $s=\frac{u}{2}$ and $\text{d}s=\frac{1}{2}\space\text{d}u$:

$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\text{C}\pm x\Longleftrightarrow$$

Now, use:
$$\int\frac{1}{\sqrt{1-x^2}}\space\text{d}x=\arcsin(x)+\text{C}$$

$$\arcsin\left(s\right)=\text{C}\pm x\Longleftrightarrow\arcsin\left(\frac{u}{2}\right)=\text{C}\pm x\Longleftrightarrow\arcsin\left(\frac{y(x)}{2}\right)=\text{C}\pm x$$
