Question

y² dx + (x² − xy + y²) dy = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {y}^{2}dx+\left(x^2-xy+y^2\right)dy=0 \\ \frac{dy}{dx}=\frac{-y^2}{x^2-xy+y^2} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{-v^2{x}^2}{x^2-v{x}^2+v^2{x}^2} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{-v^2}{1-v+v^2} \\ \Rightarrow x\frac{dv}{dx}=\frac{-v^2}{1-v+v^2}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{-v-v^3}{1-v+v^2} \\ \Rightarrow \frac{1-v+v^2}{v+v^3}dv=-\frac{1}{x}dx \\ \Rightarrow \frac{1+v^2-v}{v\left(1+v^2\right)}dv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1+v^2-v}{v\left(1+v^2\right)}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1+v^2}{v\left(1+v^2\right)}dv-\int \frac{v}{v\left(1+v^2\right)}dv=-\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{v}dv-\int \frac{1}{1+v^2}dv=-\int \frac{1}{x}dx \\ \Rightarrow \log \left|v\right|-\tan {}^{-1}\left|v\right|=-\log \left|x\right|+\log C \\ \Rightarrow \log \left|\frac{vx}{C}\right|={\tan }^{-1}v \\ \Rightarrow \left|\frac{vx}{C}\right|=e^{{\tan }^{-1}v} \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow \left|y\right|=C{e}^{{\tan }^{-1}v} \\ \mathrm{Hence},\left|y\right|=\mathrm{C}{e}^{{\tan }^{-1}v}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$