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(A) The given differential equation is $\frac{dy}{dx} + \frac{x}{x^2-1}y = \frac{x^6+4x}{\sqrt{1-x^2}}$.This is a linear differential equation of the

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$27$

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(A) The given differential equation is $\frac{dy}{dx} + \frac{x}{x^2-1}y = \frac{x^6+4x}{\sqrt{1-x^2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.Integrating factor ($I$.$F$.) = $e^{\int P(x) dx} = e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = \sqrt{1-x^2}$ (since $-1 The solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx = \int (x^6+4x) dx = \frac{x^7}{7} + 2x^2 + C$.Given $f(0)=0$,we have $0 = 0 + 0 + C \Rightarrow C=0$.Thus,$f(x) = \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}}$.We need to evaluate $6 \int_{-1/2}^{1/2} f(x) dx = 6 \int_{-1/2}^{1/2} \frac{x^7/7 + 2x^2}{\sqrt{1-x^2}} dx$.Since $\frac{x^7/7}{\sqrt{1-x^2}}$ is an odd function,its integral over $[-1/2, 1/2]$ is $0$.So,the expression becomes $6 \int_{-1/2}^{1/2} \frac{2x^2}{\sqrt{1-x^2}} dx = 24 \int_0^{1/2} \frac{x^2}{\sqrt{1-x^2}} dx$.Let $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x=0, 	heta=0$; when $x=1/2, 	heta=\pi/6$.Integral = $24 \int_0^{\pi/6} \frac{\sin^2 	heta}{\cos 	heta} \cos 	heta d	heta = 24 \int_0^{\pi/6} \sin^2 	heta d	heta = 24 \int_0^{\pi/6} \frac{1-\cos 2	heta}{2} d	heta = 12 [	heta - \frac{\sin 2	heta}{2}]_0^{\pi/6} = 12(\frac{\pi}{6} - \frac{\sqrt{3}}{4}) = 2\pi - 3\sqrt{3}$.Comparing with $2\pi - \alpha$,we get $\alpha = 3\sqrt{3}$.Therefore,$\alpha^2 = (3\sqrt{3})^2 = 27$.

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

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