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

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$\frac{1}{2}(1-e^{\pi/2})$

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(B) The given differential equation is $(1+x^2) \frac{dy}{dx} + y = e^{	an^{-1} x}$.Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{	an^{-1} x}}{1+x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{1+x^2}$ and $Q = \frac{e^{	an^{-1} x}}{1+x^2}$.The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} = e^{	an^{-1} x}$.The solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \cdot e^{	an^{-1} x} = \int \frac{e^{	an^{-1} x}}{1+x^2} \cdot e^{	an^{-1} x} dx$.Let $	an^{-1} x = z$,then $\frac{1}{1+x^2} dx = dz$.$y \cdot e^{	an^{-1} x} = \int e^{2z} dz = \frac{e^{2z}}{2} + C = \frac{e^{2	an^{-1} x}}{2} + C$.Given $y(1) = 0$,so $0 \cdot e^{	an^{-1}(1)} = \frac{e^{2	an^{-1}(1)}}{2} + C \Rightarrow 0 = \frac{e^{\pi/2}}{2} + C \Rightarrow C = -\frac{e^{\pi/2}}{2}$.Thus,$y \cdot e^{	an^{-1} x} = \frac{e^{2	an^{-1} x}}{2} - \frac{e^{\pi/2}}{2}$.For $x=0$,$y \cdot e^{	an^{-1}(0)} = \frac{e^{2	an^{-1}(0)}}{2} - \frac{e^{\pi/2}}{2} \Rightarrow y \cdot 1 = \frac{1}{2} - \frac{e^{\pi/2}}{2}$.Therefore,$y(0) = \frac{1}{2}(1 - e^{\pi/2})$.

Let $y=y(x)$ be the solution of the differential equation $(1+x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x}$,with $y(1)=0$. Then $y(0)$ is

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