Let y = y ( x ) be the solution of the differential equation (1 - x...

Given,

$\left(1 - {x}^{2}\right)dy = \left[xy + \left({x}^{3} + 2\right)\sqrt{3\left(1 - {x}^{2}\right)}\right]dx$

$⇒ \dfrac{dy}{dx} - \dfrac{xy}{1 - {x}^{2}} = \dfrac{\left({x}^{3} + 2\right)\sqrt{3\left(1 - {x}^{2}\right)}}{1 - {x}^{2}}$

The above equation is a linear differential equation,

So, $IF = {e}^{- ∫ \dfrac{x}{1 - {x}^{2}}dx} = {e}^{\dfrac{1}{2}\ln \left(1 - {x}^{2}\right)} = \sqrt{1 - {x}^{2}}$

Now, the solution of the differential equation is given by,

$y\sqrt{1 - {x}^{2}} = ∫ \dfrac{\left({x}^{3} + 2\right)\sqrt{3\left(1 - {x}^{2}\right)}}{1 - {x}^{2}} \times \sqrt{\left(1 - {x}^{2}\right)}dx$

$⇒ y\sqrt{1 - {x}^{2}} = \sqrt{3} ∫ \left({x}^{3} + 2\right)dx$

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

Now, using $y ( 0 ) = 0$ we get, $c = 0$

Now, finding $y\left(\dfrac{1}{2}\right)$

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

$⇒ y = \left(\dfrac{1}{32} + 2\right)$

$⇒ y\left(\dfrac{1}{2}\right) = \dfrac{65}{32} = \dfrac{m}{n}$

Hence, $m + n = 97$