Question

Solution of the given D.E $y^{2}dx+(x^2-xy+y^2)dy=0$ is

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

Answer 1

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

$y^{2}dx+(x^2-xy+y^2)dy=0\Rightarrow (x^2-xy+y^2)\frac{dy}{dx}+y^2=0$
Substituting $y=xv\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$(x^2-x\left(xv\right)+x^2{v}^2)(v+x\frac{dv}{dx})+x^{2}v=0\Rightarrow x^2(x\frac{dv}{dx}+v^3+x{v}^2\frac{dv}{dx}+v-xv\frac{dv}{dx})=0\Rightarrow \frac{dv}{dx}=\frac{-v^3-v}{x(v^2-v+1)}\Rightarrow \frac{dv}{dx}\times \frac{v^2-v+1}{-v^3-v}=\frac{1}{x}$
Integrating both sides, we get
$\int dv\left(\frac{v^2-v+1}{-v^3-v}\right)=\int \frac{1}{x}dx\Rightarrow ta{n}^{-1}v-\log v=\log x+c\Rightarrow ta{n}^{-1}v-\log v-\log x+c=0\Rightarrow ta{n}^{-1}\frac{y}{x}-\log y+c=0$