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