The integrating factor of the differential eq | vedclass

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Question

(D) The given differential equation is $(	an ^{-1} y - x) dy = (1+y^2) dx$. Rearranging the terms,we get $\frac{dx}{dy} = \frac{	an ^{-1} y - x}{1+y

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

Vedclass · Answer

(D) The given differential equation is $(	an ^{-1} y - x) dy = (1+y^2) dx$. Rearranging the terms,we get $\frac{dx}{dy} = \frac{	an ^{-1} y - x}{1+y^2}$. This can be written as $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{	an ^{-1} y}{1+y^2}$. 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 $Q(y) = \frac{	an ^{-1} y}{1+y^2}$. The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy}$. $IF = e^{\int \frac{1}{1+y^2} dy} = e^{	an ^{-1} y}$. Thus,the correct option is $D$.

The integrating factor of the differential equation $(\tan ^{-1} y-x) dy = (1+y^2) dx$ is . . . . . . .

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