A) \[{{x}^{2}}-xy+{{y}^{2}}=c\]
B) \[{{x}^{2}}-xy-{{y}^{2}}=c\]
C) \[{{x}^{2}}+xy-{{y}^{2}}=c\]
D) \[{{x}^{2}}+x{{y}^{2}}=c\]
Correct Answer: B
Solution :
Given equation is \[\frac{dy}{dx}=\frac{2x-y}{x+2y}\] ?..(i) Put \[y=vx\,\Rightarrow \frac{dy}{dx}=v+\frac{x\,dv}{dx}\] \[\therefore \] \[v+x\frac{dv}{dx}=\frac{2-v}{1+2v}\] \[\Rightarrow \] \[\frac{x\,dv}{dx}=\frac{2-v-v(1+2v)}{1+2v}\] \[\Rightarrow \] \[\int{\frac{1+2v}{2(1-v-{{v}^{2}})}}\,dv=\int{\frac{1}{x}\,dx}\] \[\Rightarrow \] \[\log \,k-\frac{1}{2}\,\log (1-v-{{v}^{2}})=\log x\] \[\Rightarrow \] \[2\,\log k-\log (1-v-{{v}^{2}})=2\,\log x\] \[\Rightarrow \] \[\log \,c=\log [{{x}^{2}}(1-v-{{v}^{2}})]\] \[\Rightarrow \] \[c={{x}^{2}}\left( 1-\frac{y}{x}-\frac{{{y}^{2}}}{{{x}^{2}}} \right)\] \[\Rightarrow \] \[{{x}^{2}}-xy-{{y}^{2}}=c\]
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