The solution of the differential equation (dy)/(dx) = - ( x ^ 2 + 3...

Given,

Differential equation,

$\dfrac{dy}{dx} = - \left(\dfrac{{x}^{2} + 3{y}^{2}}{3{x}^{2} + {y}^{2}}\right) \text{ and } y\left(1\right) = 0$

Now let $y = vx$

$v + x\dfrac{dv}{dx} = \dfrac{dy}{dx}$

$⇒ v + x\dfrac{dv}{dx} = - \left(\dfrac{1 + 3{v}^{2}}{3 + {v}^{2}}\right)$

$⇒ x\dfrac{dv}{dx} = - \dfrac{{\left(v + 1\right)}^{3}}{3 + {v}^{2}}$

$⇒ \dfrac{\left(3 + {v}^{2}\right)dv}{{\left(v + 1\right)}^{3}} + \dfrac{dx}{x} = 0$

$⇒ ∫ \dfrac{4dv}{{\left(v + 1\right)}^{3}} + ∫ \dfrac{dv}{v + 1} - ∫ \dfrac{2dv}{{\left(v + 1\right)}^{2}} + ∫ \dfrac{dx}{x} = 0$

$⇒ \dfrac{- 2}{{\left(v + 1\right)}^{2}} + \ln \left|v + 1\right| + \dfrac{2}{v + 1} + \ln \left|x\right| = c$

$⇒ \dfrac{- 2{x}^{2}}{{\left(x + y\right)}^{2}} + \ln \left|\dfrac{x + y}{x}\right| + \dfrac{2x}{x + y} + \ln \left|x\right| = c$

$⇒ \dfrac{2xy}{{\left(x + y\right)}^{2}} + \ln \left|x + y\right| = c$

$∵ y\left(1\right) = 0$

$∴ c = 0$, as $x = 1, y = 0$

$⇒ \dfrac{2xy}{{\left(x + y\right)}^{2}} + \ln \left|x + y\right| = 0$