Question

The general solution of $y^{2}dx+\left(x^2-xy+y^2\right)dy=0$ 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
$y^{2}dx+\left(x^2-xy+y^2\right)dy=0.$
$\Rightarrow \frac{dx}{dy}=-\left(\frac{x^2-xy+y^2}{y^2}\right)....$(i)
$=-{\left(\frac{x}{y}\right)}^2+\left(\frac{x}{y}\right)-1$
Now, put $\frac{x}{y}=v$
$\Rightarrow x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$
$\therefore v+y\frac{dv}{dy}=-v^2+v-1$
$\Rightarrow y\frac{dv}{dy}=-v^2+v-1-v=-\left(v^2+1\right)$
$\Rightarrow y\frac{dv}{dy}=-\left(v^2+1\right)$
$\Rightarrow \frac{dv}{v^2+1}=-\frac{dy}{y}$
On integrating both sides, we get
$\int \frac{dv}{v^2+1}=-\int \frac{dy}{y}$
$\Rightarrow {\tan }^{-1}v=-\log |y|+C$
$\Rightarrow {\tan }^{-1}\left(\frac{x}{y}\right)=-\log |y|+C$
$\Rightarrow {\tan }^{-1}\left(\frac{x}{y}\right)+\log |y|=C$