DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

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

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

$\therefore \frac{dx}{x} = \frac{1 + 2v^{2}}{(2v^{2} + 3)v}$

$= -\bigg[ \frac{2v^{2} + 3 - 2}{v (2v^{2} +3}\bigg] dv = \frac{-dv}{v} + \frac{2dv}{v (2v^{2} + 3)}$ 

$\therefore \frac{dx}{v} = \frac{-dv}{v} + \bigg[ \frac{A}{v} + \frac{Bv + C}{2v^{2} +3}\bigg]$

$A = \frac{2}{3}, B = -\frac{4}{3}, C = 0$

$\therefore \frac{dx}{x} = -\frac{dv}{v} + \frac{2}{3} \frac{dv}{v} -\frac{1}{3} \frac{4vdv}{2v^{2} + 3}$

Intergrating, we get

$\log x = -\log v + \frac{2}{3} \log v -\frac{1}{3} \log(2v^{2} + 3) + \log \text{c}$

$= -\frac{1}{3} \log v -\frac{1}{3} \log (2v^{2} + 3) + \log c$

$\log x^-3 = \log cv(2v^{2} + 3)$ or

$\frac{1}{x^{3}} = c \frac{y}{x} \bigg(\frac{2y^{2}}{x^{2}} + 3 \bigg)$ or

$= cy \bigg(\frac{2y^{2}}{x^{3}} + \frac{3x^{2}}{x^{3}}\bigg)$ or

$cy (2y^{2} + 3x^{2} = 1$