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(C) The given differential equation is $(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1$.Dividing by $(x^2 + 1)^2$,we get $\frac{dy}{dx} + \frac{2x}{x^2

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$1/16$

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(C) The given differential equation is $(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1$.Dividing by $(x^2 + 1)^2$,we get $\frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{1}{(x^2 + 1)^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{x^2 + 1}$ and $Q(x) = \frac{1}{(x^2 + 1)^2}$.The integrating factor $I.F. = e^{\int P(x) dx} = e^{\int \frac{2x}{x^2 + 1} dx} = e^{\ln(x^2 + 1)} = x^2 + 1$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y(x^2 + 1) = \int \frac{1}{(x^2 + 1)^2} \cdot (x^2 + 1) dx + C = \int \frac{1}{x^2 + 1} dx + C$.$y(x^2 + 1) = 	an^{-1}(x) + C$.Given $y(0) = 0$,we have $0(0^2 + 1) = 	an^{-1}(0) + C$,which implies $C = 0$.Thus,$y(x) = \frac{	an^{-1}(x)}{x^2 + 1}$.We are given $\sqrt{a} y(1) = \frac{\pi}{32}$.$y(1) = \frac{	an^{-1}(1)}{1^2 + 1} = \frac{\pi/4}{2} = \frac{\pi}{8}$.Substituting this into the equation: $\sqrt{a} \cdot \frac{\pi}{8} = \frac{\pi}{32}$.$\sqrt{a} = \frac{8}{32} = \frac{1}{4}$.Therefore,$a = (1/4)^2 = 1/16$.

Let $y = y(x)$ be the solution of the differential equation $(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1$ such that $y(0) = 0$. If $\sqrt{a} y(1) = \frac{\pi}{32}$,then the value of $a$ is

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