Question

Let $y = y(x)$ be the solution of the differential equation $(1 - x^2) \, dy = (xy + (x^3 + 2) \sqrt{1 - x^2}) \, dx \quad -1 < x < 1$, 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_______.

Answer 1

$(1 - x^2) \, dy = (xy + (x^3 + 2) \sqrt{1 - x^2}) \, dx$

$∴$ $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{x^3 + 3}{\sqrt{1 - x^2}}$

∴$$ I.F.= $e^{\int_{-\frac{x}{1-x^2}}dx} = \sqrt{1 - x^2}$
Solution is
$y \cdot \sqrt{1 - x^2} = \int (x^3 + 3) \, dx$

$y \cdot \sqrt{1 - x^2} = \frac{x^4}{4} + 3x + c$
$∵y(0)=0⇒c=0$

$∴$ $y(x) = \frac{x^4 + 12x}{4\sqrt{1 - x^2}}$

$∴$ $\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{1 - x^2} \, y(x) \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x^4 + 12x}{4} \, dx$

$=$ $\int_{0}^{\frac{1}{2}} \frac{x^4}{2} \, dx$

∴$$ k=\frac{1}{320}
$∴$ $k^{−1}=320$