Question

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

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

Answer 1

The correct option is D $\log y={\tan }^{-1}\frac{y}{x}+C$

We have $y^{2}dx+(x^2-xy+y^2)dy=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y^2}{x^2-xy+y^2}$
On putting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we get
$x\frac{dv}{dx}=\frac{-v-v^3}{v^2-v+1}$
$\Rightarrow (\frac{1}{v}-\frac{1}{v^2+1})=-\frac{dx}{x}$
On integrating on both sides we get
$\log v-{\tan }^{-1}v=-\log x+C$
$\Rightarrow \log \left(\frac{y}{x}\right)-{\tan }^{-1}\frac{y}{x}=-\log x+C$
$\Rightarrow \log y={\tan }^{-1}\frac{y}{x}+C$