Q90 / 160 · AP EAMCET 2003 · Mathematics
The general solution of $y^2 d x+\left(x^2-x y+y^2\right) d y=0$ is :
- Correct answerA$\tan ^{-1}\left(\frac{y}{x}\right)=\log y+C$
- B$2 \tan ^{-1}\left(\frac{x}{y}\right)+\log x+C=0$
- C$\log \left(y+\sqrt{x^2+y^2}\right)+\log y+C=0$
- D$\sinh ^{-1}\left(\frac{x}{y}\right)+\log y+C=0$
Solution
We have,
$y^2 d x+\left(x^2-x y+y^2\right) d y=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-y^2}{x^2-x y+y^2}$
It is a homogeneous linear differential equation
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2-v x^2+x^2 v^2}=\frac{-v^2}{v^2-v+1}$
$x \frac{d v}{d x}=\frac{-v^2-v^3+v^2-v}{v^2-v+1}=\frac{-v^3-v}{v^2-v+1}$
$\Rightarrow \quad \frac{\left(v^2-v+1\right)}{-v^3-v} d v=\frac{1}{x} d x$
$\Rightarrow \quad \frac{-\left(v^2+1\right)+v}{v\left(v^2+1\right)} d v=\frac{1}{x} d x$
$\therefore \quad-\frac{1}{v} d v+\frac{1}{v^2+1} d v=\frac{1}{x} d x$
On integrating both sides, we get
$-\log v+\tan ^{-1} v=\log x+C$
$\Rightarrow \quad \tan ^{-1} v=\log x v+C$
$\therefore \quad \tan ^{-1}\left(\frac{y}{x}\right)=\log y+C$
Hence, the correct answer is option: 'A'.