Let y=y(x) solve (1-x^2),dy = [xy+(x^3+2)√1-x^2]dx, -1

Question

Let $y=y(x)$ solve $(1-x^2)\,dy = \left[xy+(x^3+2)\sqrt{1-x^2}\right]dx$, $-1<x<1$, $y(0)=0$. If $\int_{-1/2}^{1/2}\sqrt{1-x^2}\,y(x)\,dx=k$, then $k^{-1}$ is \_\_\_\_\_\_.

Answer: 320

Accepted Answer

$$\frac{dy}{dx}-\frac{x}{1-x^2}y = \frac{x^3+3}{\sqrt{1-x^2}}\ (\text{roughly, after simplification}).$$

I.F.$=\sqrt{1-x^2}$; solving with $y(0)=0$ gives $y(x)\sqrt{1-x^2}=\dfrac{x^4}{4}+3x\cdot(\dots)$, and the requested integral (odd terms vanish) reduces to $k=\dfrac{1}{320}$, so $k^{-1}=320$.

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