The solution of the differential equation (1+ | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given differential equation: $(1+y^2) + (x - e^{	an^{-1} y}) \frac{dx}{dy} = 0$. Rearranging the terms: $(x - e^{	an^{-1} y}) \frac{dx}{dy} = -(1+y^2)$,which implies $\frac{dx}{dy} = -\frac{1+y^2}{x - e^{	an^{-1} y}}$. This is not standard. Let us rewrite as: $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{	an^{-1} y}}{1+y^2}$. 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 $Q(y) = \frac{e^{	an^{-1} y}}{1+y^2}$. Integrating Factor $(IF) = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$. The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$. Substituting the values: $x e^{	an^{-1} y} = \int \frac{e^{	an^{-1} y}}{1+y^2} \cdot e^{	an^{-1} y} dy + C$. $x e^{	an^{-1} y} = \int \frac{e^{2 	an^{-1} y}}{1+y^2} dy + C$. 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 $2$: $2 x e^{	an^{-1} y} = e^{2 	an^{-1} y} + C$.

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

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