The solution of the differential equation (1+ | vedclass

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

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$2 x e^{	an ^{-1} y}-e^{2 	an ^{-1} y}=c$

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(C) Given the differential equation: $(1+y^2)+(x-e^{	an ^{-1} y}) \frac{dy}{dx}=0$. Rearranging the equation: $(x-e^{	an ^{-1} y}) \frac{dy}{dx} = -(1+y^2)$. Dividing by $\frac{dy}{dx}$,we get $\frac{dx}{dy} = -\frac{x-e^{	an ^{-1} y}}{1+y^2} = -\frac{x}{1+y^2} + \frac{e^{	an ^{-1} y}}{1+y^2}$. 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 general solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. $x e^{	an ^{-1} y} = \int \frac{e^{	an ^{-1} y}}{1+y^2} \cdot e^{	an ^{-1} y} dy + c = \int \frac{e^{2 	an ^{-1} y}}{1+y^2} dy + c$. Let $u = 	an ^{-1} y$,then $du = \frac{1}{1+y^2} dy$. $x e^{	an ^{-1} y} = \int e^{2u} du + c = \frac{1}{2} e^{2u} + c = \frac{1}{2} e^{2 	an ^{-1} y} + c$. Multiplying by $2$,we get $2x e^{	an ^{-1} y} = e^{2 	an ^{-1} y} + 2c$. Since $2c$ is a constant,we can write the solution as $2x e^{	an ^{-1} y} - e^{2 	an ^{-1} y} = C$.

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

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