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

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$\frac{2}{e}$

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(B) The given differential equation is $(	an^{-1} y - x) dy = (1 + y^2) dx$.Rearranging the terms,we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1 + y^2}$,which can be written as $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{	an^{-1} y}{1 + y^2}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1 + y^2}$ and $Q(y) = \frac{	an^{-1} y}{1 + y^2}$.The integrating factor is $I.F. = e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{	an^{-1} y}$.The solution is $x \cdot e^{	an^{-1} y} = \int Q(y) \cdot e^{	an^{-1} y} dy + C$.Let $u = 	an^{-1} y$,then $du = \frac{1}{1 + y^2} dy$. The integral becomes $\int u e^u du = u e^u - e^u + C$.Thus,$x e^{	an^{-1} y} = (	an^{-1} y - 1) e^{	an^{-1} y} + C$.Since the curve passes through $(1, 0)$,we substitute $x = 1$ and $y = 0$: $1 \cdot e^0 = (0 - 1) e^0 + C \Rightarrow 1 = -1 + C \Rightarrow C = 2$.The equation of the curve is $x e^{	an^{-1} y} = (	an^{-1} y - 1) e^{	an^{-1} y} + 2$.For $y = 	an(1)$,$	an^{-1} y = 1$. Substituting this into the equation:$x e^1 = (1 - 1) e^1 + 2 \Rightarrow x e = 2 \Rightarrow x = \frac{2}{e}$.

If the solution curve of the differential equation $(\tan^{-1} y - x) dy = (1 + y^2) dx$ passes through the point $(1, 0)$,then the abscissa of the point on the curve whose ordinate is $\tan(1)$ is

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