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 $$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^2} y(x)dx = k$$ then $$k^{-1}$$ is equal to ______

The differential equation is $$(1 - x^2)\,dy = \bigl(xy + (x^3 + 2)\sqrt{1-x^2}\bigr)\,dx$$ with the initial condition $$y(0) = 0$$.

Rewriting in standard form gives $$\frac{dy}{dx} - \frac{x}{1-x^2}\,y = \frac{(x^3+2)\sqrt{1-x^2}}{1-x^2} = \frac{x^3+2}{\sqrt{1-x^2}}\,. $$

To find the integrating factor, we compute $$\mu = e^{\int -\frac{x}{1-x^2}\,dx}\,. $$ Substituting $$u = 1-x^2\,,\quad du = -2x\,dx$$ transforms the integral into $$\int \frac{-x}{1-x^2}\,dx = \frac{1}{2}\ln(1-x^2)\,. $$ Hence $$\mu = e^{\frac{1}{2}\ln(1-x^2)} = \sqrt{1-x^2}\,. $$

Multiplying the differential equation by $$\mu$$ yields

$$\frac{d}{dx}\bigl[y\sqrt{1-x^2}\bigr] = \sqrt{1-x^2}\,\frac{x^3+2}{\sqrt{1-x^2}} = x^3 + 2\,. $$

Integrating both sides with respect to $$x$$ gives

$$y\sqrt{1-x^2} = \frac{x^4}{4} + 2x + C\,. $$

Applying the initial condition $$y(0)=0$$ leads to $$0 = 0 + 0 + C\,, $$ so $$C = 0$$. Therefore

$$y\sqrt{1-x^2} = \frac{x^4}{4} + 2x\,. $$

We now compute

$$k = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1-x^2}\,y(x)\,dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \Bigl(\frac{x^4}{4} + 2x\Bigr)\,dx\,. $$

The term $$2x$$ is odd, so its integral over the symmetric interval $$[-\tfrac12,\tfrac12]$$ is zero. The term $$\tfrac{x^4}{4}$$ is even, so

$$k = 2\int_{0}^{\frac{1}{2}} \frac{x^4}{4}\,dx = \frac{1}{2}\Bigl[\frac{x^5}{5}\Bigr]_0^{\frac{1}{2}} = \frac{1}{2}\cdot\frac{1}{5\cdot 32} = \frac{1}{320}\,. $$

Therefore

$$k^{-1} = \mathbf{320}\,. $$