Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$, $-1

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The integrating factor is $\sqrt{1-x^2}$, so $y\sqrt{1-x^2}=\int(x^6+4x)\,dx=\frac{x^7}7+2x^2$ (with $f(0)=0$). The odd part integrates to zero over the symmetric interval, so $\int_{-1/2}^{1/2}f\,dx=4\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx=4\left(\frac\pi{12}-\frac{\sqrt3}8\right)=\frac\pi3-\frac{\sqrt3}2$. Then $6\int f=2\pi-3\sqrt3$, so $\alpha=3\sqrt3$ and $\alpha^2=27$. Answer: 27.

Q. Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$, $-1<x<1$ such that $f(0)=0$. If $6\int_{-1/2}^{1/2}f(x)\,dx=2\pi-\alpha$ then $\alpha^2$ is equal to _______ .

Explanation

Reviewed by Dr. Abubacker

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