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

$$ \Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$$

$$ \Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$$

$$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$$

So $${y \over {{{({x^2} + 4)}^3}}} = \int {{{2x} \over {{{({x^2} + 4)}^3}}}dx + c} $$

$$ \Rightarrow y = - {1 \over 2}({x^2} + 4) + c{({x^2} + 4)^3}$$

When x = 0, y = 0 gives $$c = {1 \over {32}}$$,

So, for x = 2, y = 12