Question

The general solution of $y^2 d x+\left(x^2-x y+y^2\right) d y=0$ is

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

Answer 1

Answer: (A)

The given differential equation is
$ y^2 d x+\left(x^2-x y+y^2\right) d y=0 .$
$ \Rightarrow \frac{d x}{d y}=-\left(\frac{x^2-x y+y^2}{y^2}\right) ....$(i)
$ =-\left(\frac{x}{y}\right)^2+\left(\frac{x}{y}\right)-1 $
Now, put $ \frac{x}{y}=v $
$ \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 $
$\Rightarrow y \frac{d v}{d y}=-v^2+v-1-v=-\left(v^2+1\right) $
$ \Rightarrow y \frac{d v}{d y}=-\left(v^2+1\right) $
$ \Rightarrow \frac{d v}{v^2+1}=-\frac{d y}{y}$
On integrating both sides, we get
$\int \frac{d v}{v^2+1} =-\int \frac{d y}{y} $
$\Rightarrow \tan ^{-1} v =-\log |y|+C $
$\Rightarrow \tan ^{-1}\left(\frac{x}{y}\right) =-\log |y|+C $
$\Rightarrow \tan ^{-1}\left(\frac{x}{y}\right)+\log |y| =C$