Question

Find the particular solution of the differential equation $\left(x-y\right)\frac{dy}{dx}=x+2y$, given that when x = 1, y = 0.

Answer 1

$$\begin{gathered} \left(x-y\right)\frac{dy}{dx}=x+2y \\ \Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-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}{x-vx} \\ \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{1-v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^2}{1-v} \\ \Rightarrow x\frac{dv}{dx}=\frac{1+v+v^2}{1-v} \\ \Rightarrow \frac{1-v}{1+v+v^2}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1-v}{1+v+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{1+v+v^2}dv-\frac{1}{2}\int \frac{2v+1-1}{1+v+v^2}=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{1+v+v^2}dv-\frac{1}{2}\int \frac{2v+1}{1+v+v^2}dv+\frac{1}{2}\int \frac{1}{1+v+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \frac{3}{2}\int \frac{1}{1+v+v^2}dv-\frac{1}{2}\int \frac{2v+1}{1+v+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \frac{3}{2}\int \frac{1}{1+v+v^2+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}}dv-\frac{1}{2}\int \frac{2v+1}{1+v+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \frac{3}{2}\int \frac{1}{{\left(v+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dv-\frac{1}{2}\int \frac{2v+1}{1+v+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow \sqrt{3}\tan {}^{-1}\left|\frac{v+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right|-\frac{1}{2}\log \left|1+v+v^2\right|=\log \left|x\right|+C \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{1}{2}\log \left|\frac{x^2+xy+y^2}{x^2}\right|=\log \left|x\right|+C \\ \Rightarrow \sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{1}{2}\log \left|x^2+xy+y^2\right|+\log \left|x\right|=\log \left|x\right|+C \\ \Rightarrow \sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{1}{2}\log \left|x^2+xy+y^2\right|=C.....\left(1\right) \\ \mathrm{At}x=1,y=0\left(\mathrm{Given}\right) \\ \mathrm{Putting}x=1\mathrm{and}y=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \sqrt{3}{\tan }^{-1}\left|\frac{1}{\sqrt{3}}\right|-\frac{1}{2}\log \left|1\right|=C \\ \Rightarrow C=\sqrt{3}{\tan }^{-1}\left|\frac{1}{\sqrt{3}}\right| \\ \Rightarrow C=\sqrt{3}\times \frac{\mathrm{\pi }}{6} \\ \Rightarrow C=\frac{\mathrm{\pi }}{2\sqrt{3}} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{1}{2}\log \left|x^2+xy+y^2\right|=\frac{\mathrm{\pi }}{2\sqrt{3}} \\ \Rightarrow 2\sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\log \left|x^2+xy+y^2\right|=\frac{\mathrm{\pi }}{\sqrt{3}} \\ \Rightarrow \log \left|x^2+xy+y^2\right|=2\sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{\mathrm{\pi }}{\sqrt{3}} \\ \mathrm{Hence},\log \left|x^2+xy+y^2\right|=2\sqrt{3}{\tan }^{-1}\left|\frac{2y+x}{\sqrt{3}x}\right|-\frac{\mathrm{\pi }}{\sqrt{3}}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$