Let the solution curve y = y (x) of the differential equation (4 +...

Given $\left(4 + {x}^{2}\right)dy - 2x\left({x}^{2} + 3y + 4\right)dx = 0$

\left({x}^{2} + 4\right)\dfrac{dy}{dx} = 2{x}^{3} + 6xy + 8x$</p><p>\left({x}^{2} + 4\right)\dfrac{dy}{dx} - 6xy = 2{x}^{3} + 8x$

$\dfrac{dy}{dx} - \dfrac{6x}{{x}^{2} + 4}y = \dfrac{2{x}^{3} + 8x}{{x}^{2} + 4}$

This is of the form of linear differential equation

I.F. $= {e}^{- ∫ \dfrac{6x}{{x}^{2} + 4}dx} = {e}^{- 3\log_{e}\left({x}^{2} + 4\right)}$

$= {e}^{\log_{e}{\left({x}^{2} + 4\right)}^{- 3}} = \dfrac{1}{{\left({x}^{2} + 4\right)}^{3}}$

So the solution of the differential equation will be

$y . \dfrac{1}{{\left({x}^{2} + 4\right)}^{3}} = ∫ \dfrac{2{x}^{3} + 8x}{{\left({x}^{2} + 4\right)}^{3}\left({x}^{2} + 4\right)}dx$

$\dfrac{y}{{\left({x}^{2} + 4\right)}^{3}} = ∫ \dfrac{2x\left({x}^{2} + 4\right)}{{\left({x}^{2} + 4\right)}^{3}\left({x}^{2} + 4\right)}dx$

Let ${x}^{2} + 4 = t$, $2xdx = dt$

So $\dfrac{y}{{\left({x}^{2} + 4\right)}^{3}} = ∫ \dfrac{dt}{{t}^{3}}$

$\dfrac{y}{{\left({x}^{2} + 4\right)}^{3}} = \dfrac{- 1}{2{\left({x}^{2} + 4\right)}^{2}} + C$

Since the curve passes through origin \left(0, 0\right)$</p><p>$0 = \dfrac{- 1}{2 \times 16} + C ⇒ C = \dfrac{1}{32}$</p><p>i.e.$\dfrac{y}{{\left({x}^{2} + 4\right)}^{3}} = \dfrac{- 1}{2{\left({x}^{2} + 4\right)}^{2}} + \dfrac{1}{32}$</p><p>$y = \dfrac{- \left({x}^{2} + 4\right)}{2} + \dfrac{{\left({x}^{2} + 4\right)}^{3}}{32}$</p><p>Hence$y\left(2\right) = - \dfrac{8}{2} + \dfrac{8 \times 8 \times 8}{32} = 12$