Question

The solution of the differential equation $ ({{x}^{2}}+{{y}^{2}})dx=2xy\,\,dy $ is

• 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)

Answer 1

Answer: (B)

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}}) $