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(C) Given the differential equation: $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$. Dividing both sides by $(1+y^2) dy$,we get: $\frac{dx}{dy} = \

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$\operatorname{Tan}^{-1} y - 1$

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(C) Given the differential equation: $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$. Dividing both sides by $(1+y^2) dy$,we get: $\frac{dx}{dy} = \frac{\operatorname{Tan}^{-1} y - x}{1+y^2}$. Rearranging the terms: $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\operatorname{Tan}^{-1} y}{1+y^2}$. 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 $Q(y) = \frac{\operatorname{Tan}^{-1} y}{1+y^2}$. The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\operatorname{Tan}^{-1} y}$. The general solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + c$. Substituting the values: $x \cdot e^{\operatorname{Tan}^{-1} y} = \int \frac{\operatorname{Tan}^{-1} y}{1+y^2} \cdot e^{\operatorname{Tan}^{-1} y} dy + c$. Let $u = \operatorname{Tan}^{-1} y$,then $du = \frac{1}{1+y^2} dy$. The integral becomes $\int u e^u du = u e^u - e^u + c$. So,$x \cdot e^{\operatorname{Tan}^{-1} y} = \operatorname{Tan}^{-1} y \cdot e^{\operatorname{Tan}^{-1} y} - e^{\operatorname{Tan}^{-1} y} + c$. Dividing by $e^{\operatorname{Tan}^{-1} y}$,we get $x = \operatorname{Tan}^{-1} y - 1 + c e^{-\operatorname{Tan}^{-1} y}$. Comparing this with $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$,we find $f(y) = \operatorname{Tan}^{-1} y - 1$.

If the general solution of $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$ is $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$,then $f(y) =$

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