Given the O.D.E. $2yy''=1+(y')²$ by using the transformation $y'=z, y''=z·(dz/dy)$ find the solutions.Why cases for y are not taken? By using this transformation we have:
$$\dfrac {2z\;dz}{z^2+1}=\dfrac {dy}y$$
Then by integrating the book comes to the conclusion that:
$z=±\sqrt {c_1y-1}$ , where $c_1$ is a non zero arbitrary constant.
How do we know that $c_1y-1 \ge 0$ in order to apply a root? Why doesn't it examine cases for $y$?

It's an autonomous equation.
We need to solve the ODE: $yy''=1+(y')^{2}$.
Now, treating $y$ as the independent variable, let $z=y'$ wich gives $y''=z\frac{dz}{dy}$, so $$2y\frac{dz}{dy}z=z^{2}+1$$Solving for $\frac{dz}{dy}$, we can see that $$\frac{dz}{dy}=\frac{z^{2}+1}{2yz} \overbrace{\implies}^{\div \frac{z^{2}+1}{2z}} \frac{2\frac{dz}{dy}z}{z^{2}+1}=\frac{1}{y} \implies \int \frac{2\frac{dz}{dy}z}{z^{2}+1} dy = \int \frac{1}{y} dy \implies \ln(z^{2}+1)=\ln(y)+c_{1}.$$
Solving for $z$, we can see that $$z=-\sqrt{e^{c_{1}}y-1} \quad \vee \quad z=\sqrt{e^{c_{1}}y-1}.$$
Now, sustituing back for $z=\frac{dy}{dx}$, we can see that $$\frac{dy}{dx}=-\sqrt{c_{1}y-1} \quad \vee \quad \frac{dy}{dx}=\sqrt{c_{1}y-1}.$$For $\frac{dy}{dx}=-\sqrt{c_{1}y-1}$, we can obtain that $$y=\frac{1}{c_{1}}+\frac{1}{4}(x-c_{2})^{2}c_{1}$$also for $\frac{dy}{dx}=\sqrt{c_{1}y-1}$, we can obtain that $$y_{x}=\frac{1}{c_{1}}+\frac{1}{4}(x+c_{2})^{2}c_{1}$$Finally, collecting solutions and simplify the arbitrary constanst we can see that $$\boxed{y(x)=\frac{1}{c_{1}}+\frac{1}{4}(x+c_{2})^{2}c_{1}}.$$
