Let $y = y(x)$ be the solution of the differential equation $$(1 - {x^2})dy = \left( {xy + ({x^3} + 2)\sqrt {1 - {x^2}} } \right)dx, - 1 < x < 1$$, and $y(0) = 0$. If $\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = k}$, then k^$-$1 is equal to _____________.

<p>$$\left( {1 - {x^2}} \right)dy = \left( {xy + \left( {{x^3} + 2} \right)\sqrt {1 - {x^2}} } \right)dx$$</p> <p>$\therefore$ $${{dy} \over {dx}} - {x \over {1 - {x^2}}}y = {{{x^3} + 3} \over {\sqrt {1 - {x^2}} }}$$</p> <p>$\therefore$ $I.F. = {e^{\int { - {x \over {1 - {x^2}}}dx} }} = \sqrt {1 - {x^2}}$</p> <p>Solution is</p> <p>$y.\,\sqrt {1 - {x^2}} = \int {\left( {{x^3} + 3} \right)dx}$</p> <p>$y.\,\sqrt {1 - {x^2}} = {{{x^4}} \over 4} + 3x + c$</p> <p>$\because$ $y(0) = 0 \Rightarrow c = 0$</p> <p>$\therefore$ $y(x) = {{{x^4} + 12x} \over {4\sqrt {1 - {x^2}} }}$</p> <p>$\therefore$ $$\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = \int_{{{ - 1} \over 2}}^{{1 \over 2}} {\left( {{{{x^4} + 12x} \over 4}} \right)dx = \int_0^{{1 \over 2}} {{{{x^4}} \over 2}dx} } } $$</p> <p>$\therefore$ $k = {1 \over {320}}$</p> <p>$\therefore$ $= {k^{ - 1}} = 320$</p>