Question

(x² − 2xy) dy + (x² − 3xy + 2y²) dx = 0

Answer 1

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