Question

$x\frac{dy}{dx}=x+y$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}=x+y \\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x+vx}{x} \\ \Rightarrow v+x\frac{dv}{dx}=1+v \\ \Rightarrow x\frac{dv}{dx}=1+v-v \\ \Rightarrow x\frac{dv}{dx}=1 \\ \Rightarrow dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dv=\int \frac{1}{x}dx \\ \Rightarrow v=\log \left|x\right|+C \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{y}{x}=\log \left|x\right|+C \\ \Rightarrow y=x\log \left|x\right|+Cx \\ \mathrm{Hence},y=x\log \left|x\right|+Cx\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$