Let y = y ( x ) be the solution of the differential equation 1 - x 2 d y = x y + x 3 + 2 3 1 - x 2 d x , - 1…

Given,

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

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

The above equation is a linear differential equation,

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

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

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

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

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

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

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

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

$\Rightarrow y=\left(\frac{1}{32}+2\right)$

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

Hence, $m+n=97$