Example:
\[2y \frac{d^2y}{dx^2}=(\( \frac{dy}{dx} \)^2-1\]
.
Let \[u= \frac{dy}{dx}\]
then the equation becomes \[\frac{du}{dx}=u^2-1 \rightarrow \int \frac{1}{u^2-1}du = \int 1dx \]
.\[\frac{1}{u^2-1}d = \frac{1/2}{u-1} - \frac{1/2}{u+1}\]
so the integral becomes\[\frac{du}{dx}=u^2-1 \rightarrow \int \frac{1/2}{u-1} - \frac{1/2}{u+1} du = \int 1dx \]
.\[\frac{ln(\frac{u-1}{u+1})}{2}=x+c \rightarrow \frac{u-1}{u+1}=2x+2c \rightarrow \frac{u-1}{u+1}=e^{2x+2c}=Ae^{2x} \rightarrow u= \frac{1+Ae^{2x}}{1-Ae^{2x}} \]
.Hence
\[\frac{dy}{dx}= \frac{1+Ae^{2x}}{1-Ae^{2x}}=1+\frac{2Ae^{2x}}{1-Ae^{2x}}=1-\frac{-2Ae^{2x}}{1-Ae^{2x}}\]
.Integrating again gives
\[y=x+-ln(1-Ae^{2x})+B\]
.