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

Correct Answer: 320

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

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

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

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

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

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

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

\(∴\)\( k=\frac{1}{320}\)
\(∴\) \(k^{−1}=320\)