Question

$x^2\frac{dy}{dx}=x^2+xy+y^2$

Answer 1

$$\begin{gathered} x^2\frac{dy}{dx}=x^2+xy+y^2 \\ \Rightarrow \frac{dy}{dx}=\frac{x^2+xy+y^2}{x^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x^2+x^{2}v+v^2{x}^2}{x^2} \\ \Rightarrow v+x\frac{dv}{dx}=1+v+v^2 \\ \Rightarrow x\frac{dv}{dx}=\left(1+v^2\right) \\ \Rightarrow \frac{1}{1+v^2}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{1+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow {\tan }^{-1}v=\log \left|x\right|+C \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\log \left|x\right|+\mathrm{C} \\ \mathrm{Hence},{\tan }^{-1}\left(\frac{y}{x}\right)=\log \left|x\right|+\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$