Question

x² dy + (x² − xy + y²) dx = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {x}^{2}dy+\left(x^2-xy+y^2\right)dy=0 \\ \Rightarrow x^{2}dy=\left(xy-x^2-y^2\right)dy \\ \Rightarrow \frac{dy}{dx}=\frac{xy-x^2-y^2}{x^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x^{2}v-x^2-x^2{v}^2}{x^2} \\ \Rightarrow v+x\frac{dv}{dx}=v-1-v^2 \\ \Rightarrow x\frac{dv}{dx}=-1-v^2 \\ \Rightarrow \frac{dv}{1+v^2}=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{dv}{1+v^2}dv=-\int \frac{1}{x}dx \\ \Rightarrow {\tan }^{-1}v=-\log \left|x\right|+\log C \\ \Rightarrow {\tan }^{-1}\frac{y}{x}=\log \frac{C}{x} \\ \Rightarrow e^{{\tan }^{-1}\frac{y}{x}}=\frac{C}{x} \\ \Rightarrow C=x{e}^{{\tan }^{-1}\frac{y}{x}} \end{gathered}$$