Question

$(x^2+xy)dy=(x^2+y^2)dx$ is

• A. $\log x=\log (x-y)+\frac{y}{x}+C$
• B. $\log x=2\log (x-y)+\frac{y}{x}+C$
• C. $\log x=\log (x-y)+\frac{x}{y}+C$
• D. none of these

Answer 1

Answer: (B)

$\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}$ . Put $y=vx$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}$
$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{1-v}{1+v}$
$\therefore \frac{1+v}{1-v}dv=\frac{dx}{x}$
$\Rightarrow \left(-1+\frac{2}{1-v}\right)dv=\frac{dx}{x}$
$\Rightarrow -v-2log\left(1-v\right)+C=logx$
$\Rightarrow -\frac{y}{x}-2log(1-\frac{y}{x})+C=logx$
$\Rightarrow -\frac{y}{x}-2log+\frac{y}{x}+2log\left(\frac{x-y}{x}\right)=0$
$\frac{y}{x}+2log\left(x-y\right)+C=logx$