Question

When $y = vx$ , $y$ and $x$ are variables, the differential equation $\frac{dy}{dx}=\frac{2xy}{x^{2}-y^{2}}$ reduces to

• A. $\frac{1-v^{2}}{v+v^{3}}dv=\frac{2}{x}dx$
• B. $\frac{1-v^{2}}{v+v^{3}}dv=\frac{1}{x}dx$
• C. $\frac{1+v^{2}}{v+v^{3}}dv=\frac{1}{x}dx$
• D. $\frac{1-v^{3}}{v+v^{3}}dv=\frac{1}{x}dx$

Answer 1

Answer: (B)

$\frac{1-v^{2}}{v+v^{3}}dv=\frac{1}{x}dx$

Answer 2

We have, $\frac{dy}{dx} = \frac{2xy}{x^2 - y^2}\,\,\,...(i)$
Put $y = vx$
$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$
$\therefore$ Equation $(i)$ becomes,
$v + x \frac{dv}{dx} = \frac{2x^2v}{x^2 - v^2 x^2}$
$\Rightarrow x \frac{dv}{dx} = \frac{v + v^3}{1-v^2}$
$\Rightarrow \left(\frac{1-v^{2}}{v + v^{3}}\right) dv = \frac{dx}{x}$