Question

Solution of $y^{2}dx+(x^2-xy+y^2)dy=0$

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

Answer 1

The correct option is A $\tan {}^{-1}\left(\frac{x}{y}\right)+\log y+c=0$

$y^{2}dx+\left(x^2-xy+42\right)dy=0$

$\frac{dx}{dy}=\frac{-\left(x^2-xy+y^2\right)}{y^2}$

$\frac{dx}{dy}=-\frac{x^2}{y^2}+\frac{x}{y}-1$

Put $v=x/y⟹x=vy$

$dx/dy=v+ydv/dy$

$\Rightarrow v+ydv/dy=-v^2+v-1$

$\Rightarrow ydv/dy=-\left(v^2+1\right)$

$\frac{dv}{v^2+1}=-\frac{1}{y}dy$

Integrating both side

$\Rightarrow {\tan }^{-1}v=-\log y+c$

$\Rightarrow {\tan }^{-1}x/y=-\log y+cc=-c$

$\Rightarrow {\tan }^{-1}\left(x/y\right)+\log y+c=0$