Question

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

Answer 1

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