The solution of (1+y^2)+(x-e^{tan ^{-1} y}) f | vedclass

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Question

(A) 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} = -

Vedclass · Accepted Answer

$2x e^{	an ^{-1} y}=e^{2 	an ^{-1} y}+k$,where $k$ is the constant of integration

Vedclass · Answer

(A) Given the differential equation: $(1+y^2)+(x-e^{	an ^{-1} y}) \frac{dy}{dx}=0$. Rearranging the equation: $(x-e^{	an ^{-1} y}) \frac{dy}{dx} = -(1+y^2)$. Taking the reciprocal: $\frac{dx}{dy} = -\frac{x-e^{	an ^{-1} y}}{1+y^2} = -\frac{x}{1+y^2} + \frac{e^{	an ^{-1} y}}{1+y^2}$. 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 $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an ^{-1} y}$. The solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + k$. $x \cdot e^{	an ^{-1} y} = \int \frac{e^{	an ^{-1} y}}{1+y^2} \cdot e^{	an ^{-1} y} dy + k$. Let $u = 	an ^{-1} y$,then $du = \frac{1}{1+y^2} dy$. $x \cdot e^{	an ^{-1} y} = \int e^{2u} du + k = \frac{1}{2} e^{2u} + k = \frac{1}{2} e^{2 	an ^{-1} y} + k$. Multiplying by $2$: $2x e^{	an ^{-1} y} = e^{2 	an ^{-1} y} + 2k$. Since $2k$ is also a constant,we can write it as $k$. Thus,$2x e^{	an ^{-1} y} = e^{2 	an ^{-1} y} + k$.

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

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