Question

The general solution of $ {{y}^{2}}dx+({{x}^{2}}-xy+{{y}^{2}})dy=0 $ is

• A. $ {{\tan }^{-1}}\left( \frac{x}{y} \right)+\log y+c=0 $
• B. $ 2{{\tan }^{-1}}\left( \frac{x}{y} \right)+\log x+c=0 $
• C. $ \log (y+\sqrt{{{x}^{2}}+{{y}^{2}}})+\log y+c=0 $
• D. $ {{\sin }^{-1}}\left( \frac{x}{y} \right)+\log y+c=0 $

Answer 1

Answer: (A)

Given that, $y^{2} d x+\left(x^{2}-x y+y^{2}\right) d y=0 \Rightarrow $
$d x+\frac{x^{2}-x y+y^{2}}{y^{2}} d y=0 \Rightarrow \frac{d x}{d y}+\left(\frac{x}{y}\right)^{2}-\left(\frac{x}{y}\right)+1=0$
Let $v=\frac{x}{y} \Rightarrow x=v y \Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y} \therefore $
$v+y \frac{d v}{d y}+v^{2}-v+1=0 \Rightarrow y \frac{d y}{d x}=-\left(v^{2}+1\right) \Rightarrow $
$\frac{d v}{v^{2}+1}+\frac{d y}{y}=0$
On integrating, we get
$\tan ^{-1} v+\log y+c=0 \Rightarrow \tan ^{-1} \frac{x}{y}+\log y+c=0$