An integrating factor for the differential eq | vedclass

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Question

(B) Given the differential equation: $(1 + y^2)dx - (	an^{-1} y - x)dy = 0$ Rearranging the terms to form a linear differential equation in $x$: $(1

Vedclass · Accepted Answer

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

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(B) Given the differential equation: $(1 + y^2)dx - (	an^{-1} y - x)dy = 0$ Rearranging the terms to form a linear differential equation in $x$: $(1 + y^2)dx = (	an^{-1} y - x)dy$ $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1 + y^2}$ $\frac{dx}{dy} = \frac{	an^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$ Adding $\frac{x}{1 + y^2}$ to both sides: $\frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{	an^{-1} y}{1 + y^2}$ This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{1 + y^2}$ and $Q = \frac{	an^{-1} y}{1 + y^2}$. The integrating factor $(I.F.)$ is given by: $I.F. = e^{\int P dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{	an^{-1} y}$.

An integrating factor for the differential equation $(1 + y^2)dx - (\tan^{-1} y - x)dy = 0$ is:

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