A) \[{{x}^{2}}+{{y}^{2}}=cy\]
B) \[c({{x}^{2}}-{{y}^{2}})=x\]
C) \[{{x}^{2}}-{{y}^{2}}=cy\]
D) \[{{x}^{2}}+{{y}^{2}}=cx\] (here c is an arbitrary constant)
Correct Answer: B
Solution :
Given differential equation can be rewritten as \[\frac{dy}{dx}=\frac{{{x}^{2}}+{{y}^{2}}}{2xy}\] Put \[y=vx\] \[\Rightarrow \] \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] Then, given differential equation becomes \[v+x\frac{dv}{dx}=\frac{{{x}^{2}}+{{v}^{2}}{{x}^{2}}}{2xvx}\] \[\Rightarrow \] \[x\frac{dv}{dx}=\frac{1+{{v}^{2}}}{2v}-v\] \[\Rightarrow \] \[x\frac{dv}{dx}=\frac{1-{{v}^{2}}}{2v}\] \[\Rightarrow \] \[\frac{2v}{1-{{v}^{2}}}\,\,\,dv=\frac{dx}{x}\] On integrating, we get \[-\log \,\,(1-{{v}^{2}})=\log \,x+\log \,c\] \[\Rightarrow \] \[\log {{(1-{{v}^{2}})}^{-1}}=\log \,xc\] \[\Rightarrow \] \[{{\left( 1-\frac{{{y}^{2}}}{{{x}^{2}}} \right)}^{-1}}=xc\] \[\Rightarrow \] \[{{\left( \frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}} \right)}^{-1}}=xc\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{{{x}^{2}}-{{y}^{2}}}=xc\] \[\Rightarrow \] \[x=c({{x}^{2}}-{{y}^{2}})\]
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