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