\[y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2=1\]
let \[\frac{dy}{dx}=p\]
then \[\frac{d^2y}{dx^2}= \frac{dp}{dx}= \frac{dy}{dx} \frac{dp}{dy}=p \frac{dy}{dx}\]
and the equation becomes \[yp \frac{dp}{dy}+p^2=1\]
.Separation of variables gives
\[\frac{p}{1-p^2}dp= \frac{1}{y} dy\]
\[\int \frac{p}{1-p^2}dp= \int \frac{1}{y} dy\]
\[- \frac{1}{2} ln(1-p^2)= lny+c= \]
\[ln(1-p^2)=-2lny-2c=ln(\frac{1}{y^2})-2c\]
\[1-p^2=e^{ln(\frac{1}{y^2})-2c}=e^{-2c} \frac{1}{y^2}\]
\[p=\sqrt{1-e^{-2c} \frac{1}{y^2}}=\sqrt{y^2-A}{y}\]
\[p= \frac{dy}{dx}\]
so \[\frac{dy}{dx}=\sqrt{y^2-A}{y}\]
Separating variables again
\[\frac{y}{\sqrt{y^2-A}}dy= dx\]
.Integration gives
\[\sqrt{y^2-A}=x+B \rightarrow y^2=(x+B)^2+A \rightarrow y=\sqrt{(x+B)^2+A}\]
.