The integrating factor of the differential eq | vedclass

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Question

(A) The given differential equation is $(1 + t^2) + (x - e^{	an^{-1} t}) \frac{dt}{dx} = 0$.Rearranging the terms, we get $(x - e^{	an^{-1} t}) \fra

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$e^{	an^{-1} t}$

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(A) The given differential equation is $(1 + t^2) + (x - e^{	an^{-1} t}) \frac{dt}{dx} = 0$.Rearranging the terms, we get $(x - e^{	an^{-1} t}) \frac{dt}{dx} = -(1 + t^2)$.Taking the reciprocal, $\frac{dx}{dt} = -\frac{x - e^{	an^{-1} t}}{1 + t^2} = -\frac{x}{1 + t^2} + \frac{e^{	an^{-1} t}}{1 + t^2}$.This can be written as $\frac{dx}{dt} + \frac{1}{1 + t^2} x = \frac{e^{	an^{-1} t}}{1 + t^2}$.This is a linear differential equation of the form $\frac{dx}{dt} + P(t)x = Q(t)$, where $P(t) = \frac{1}{1 + t^2}$.The integrating factor $(IF)$ is given by $e^{\int P(t) dt} = e^{\int \frac{1}{1 + t^2} dt} = e^{	an^{-1} t}$.

The integrating factor of the differential equation $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$ is

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