Question

The solution of the differential equation $ ({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}+x{{y}^{2}}=0 $ is

• A. $ \log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+C $
• B. $ \log \left( \frac{y}{x} \right)=\frac{1}{x}+\frac{1}{y}+C $
• C. $ \log (xy)=\frac{1}{x}+\frac{1}{y}+C $
• D. $ \log (xy)+\frac{1}{x}+\frac{1}{y}=C $

Answer 1

Answer: (A)

Given differential equation is
$ ({{x}^{2}}-y{{x}^{2}})\frac{dy}{dx}+{{y}^{2}}=x{{y}^{2}}=0 $
$ \Rightarrow $ $ \frac{1-y}{{{y}^{2}}}dy+\frac{1+x}{{{x}^{2}}}dx=0 $
$ \Rightarrow $ $ \left( \frac{1}{{{y}^{2}}}-\frac{1}{y} \right)dy+\left( \frac{1}{{{x}^{2}}}+\frac{1}{x} \right)dx=0 $
On integrating both sides, we get the required solution
$ \frac{-2}{y}-\log y-\frac{2}{x}+\log x=C $
$ \Rightarrow $ $ \log \left( \frac{x}{y} \right)=\frac{1}{x}+\frac{1}{y}+C $