Question

The general solution of $ \frac{dy}{dx}=\frac{2x-y}{x+2y} $ is

• A. $ {{x}^{2}}-xy+{{y}^{2}}=c $
• B. $ {{x}^{2}}-xy-{{y}^{2}}=c $
• C. $ {{x}^{2}}+xy-{{y}^{2}}=c $
• D. $ {{x}^{2}}+x{{y}^{2}}=c $

Answer 1

Answer: (B)

Given equation is $ \frac{dy}{dx}=\frac{2x-y}{x+2y} $ ..(i) Put $ y=vx\,\Rightarrow \frac{dy}{dx}=v+\frac{x\,dv}{dx} $
$ \therefore $ $ v+x\frac{dv}{dx}=\frac{2-v}{1+2v} $
$ \Rightarrow $ $ \frac{x\,dv}{dx}=\frac{2-v-v(1+2v)}{1+2v} $
$ \Rightarrow $ $ \int{\frac{1+2v}{2(1-v-{{v}^{2}})}}\,dv=\int{\frac{1}{x}\,dx} $
$ \Rightarrow $ $ \log \,k-\frac{1}{2}\,\log (1-v-{{v}^{2}})=\log x $
$ \Rightarrow $ $ 2\,\log k-\log (1-v-{{v}^{2}})=2\,\log x $
$ \Rightarrow $ $ \log \,c=\log [{{x}^{2}}(1-v-{{v}^{2}})] $
$ \Rightarrow $ $ c={{x}^{2}}\left( 1-\frac{y}{x}-\frac{{{y}^{2}}}{{{x}^{2}}} \right) $
$ \Rightarrow $ $ {{x}^{2}}-xy-{{y}^{2}}=c $