How to solve the ODE:
$yy''+(y')^2=x$
How to solve the ODE:
$yy''+(y')^2=x$
Integrating both sides with respect to $x$ gives \begin{equation*} -\frac{x^2}{2}+y'y=C_1 \end{equation*} where we have used the reverse product rule and $C_1$ is a constant. Solve for $y'$ and multiply both sides by $2y$ to get \begin{equation*} 2y'y=x^2+2C_1. \end{equation*} Integrate both sides with respect to $x$ to get \begin{equation*} y^2=\frac{x^3}{3}+2C_1x+C_2 \end{equation*} where $C_2$ is another constant. Take the square root. $~_{\square}$