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, $y^{2}dx+\left(x^2-xy+y^2\right)dy=0\Rightarrow$
$dx+\frac{x^2-xy+y^2}{y^2}dy=0\Rightarrow \frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2-\left(\frac{x}{y}\right)+1=0$
Let $v=\frac{x}{y}\Rightarrow x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}\therefore$
$v+y\frac{dv}{dy}+v^2-v+1=0\Rightarrow y\frac{dy}{dx}=-\left(v^2+1\right)\Rightarrow$
$\frac{dv}{v^2+1}+\frac{dy}{y}=0$
On integrating, we get
${\tan }^{-1}v+\log y+c=0\Rightarrow {\tan }^{-1}\frac{x}{y}+\log y+c=0$