The solution of the differential equation (1 | vedclass

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(B) Given the differential equation: $(1 + y^2) + (x - e^{	an^{-1}y}) \frac{dy}{dx} = 0$. Rearranging the equation to solve for $\frac{dx}{dy}$: $(1

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$2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$

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(B) Given the differential equation: $(1 + y^2) + (x - e^{	an^{-1}y}) \frac{dy}{dx} = 0$. Rearranging the equation to solve for $\frac{dx}{dy}$: $(1 + y^2) \frac{dx}{dy} + x = e^{	an^{-1}y}$. Dividing by $(1 + y^2)$: $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{	an^{-1}y}}{1 + y^2}$. 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 $Q(y) = \frac{e^{	an^{-1}y}}{1 + y^2}$. The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{	an^{-1}y}$. The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + k$. $x e^{	an^{-1}y} = \int \frac{e^{	an^{-1}y}}{1 + y^2} \cdot e^{	an^{-1}y} dy + k$. Let $u = 	an^{-1}y$,then $du = \frac{1}{1 + y^2} dy$. $x e^{	an^{-1}y} = \int e^{2u} du + k = \frac{e^{2u}}{2} + k = \frac{e^{2	an^{-1}y}}{2} + k$. Multiplying by $2$: $2xe^{	an^{-1}y} = e^{2	an^{-1}y} + k$.

The solution of the differential equation $(1 + y^2) + (x - e^{\tan^{-1}y}) \frac{dy}{dx} = 0$ is

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