# Solve the following. $\large\frac{dy}{dx}$$+xy\;=\;x ## 1 Answer Toolbox: • Linear Differential equation. • This is of the form \large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
• The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$\large\frac{dy}{dx}$$+xy=x P(x)=x\;Q(x)=x e^{ \large \int pdx}=e^{\large\int xdx} \qquad=e^{\large\frac{x^2}{2}} The solution is ye^{\large\frac{x^2}{2}}=\int xe^{\Large\frac{x^2}{2}}dx+c ye^{\large\frac{x^2}{2}}=e^{\large\frac{x^2}{2}}$$+c$