Question

(x + 2y) dx − (2x − y) dy = 0

Answer 1

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