The function y=f(x) is the solution of the di | vedclass

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q

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$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}$.Integrating factor ($I$.$F$.) $= e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.The general solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + c = \int (x^4+2x) dx + c = \frac{x^5}{5} + x^2 + c$.Given $f(0)=0$,we have $0 \cdot 1 = 0 + 0 + c$,so $c=0$.Thus,$f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$.We need to evaluate $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$. Since $f(x) = \frac{x^5/5}{\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$,the first part is an odd function,so its integral over the symmetric interval is $0$.Thus,$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$.Let $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x=0, 	heta=0$; when $x=\frac{\sqrt{3}}{2}, 	heta=\frac{\pi}{3}$.$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 	heta}{\cos 	heta} \cos 	heta d	heta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 	heta d	heta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2	heta) d	heta$.$I = [	heta - \frac{1}{2} \sin 2	heta]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{1}{2} \sin(\frac{2\pi}{3}) = \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

The function $y=f(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$ in $(-1,1)$ satisfying $f(0)=0$. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$ is

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