Question

On substituting $y=vx$ the equation
$\frac{dy}{dx}=\frac{2xy-y^2}{2xy-x^2}$ reduces to

• A. $x(2v-1)dv=3v(v-1)dx$
• B. $x(2v-1)dv=3v(1-v)dx$
• C. $x(1-2v)dv=(v^2-2v)dx$
• D. $x(1-2v)dv=(2v-v^2)dx$

Answer 1

Answer: (B)

$x(2v-1)dv=3v(1-v)dx$

$y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

$\Rightarrow x(2v-1)dv=3v(1-v)dx$