The integrating factor of the differential eq | vedclass

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Question

(A) Given the differential equation: $(	an ^{-1} y - x) dy = (1 + y^2) dx$ Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$: $\frac{d

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

Vedclass · Answer

(A) Given the differential equation: $(	an ^{-1} y - x) dy = (1 + y^2) dx$ Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$: $\frac{dx}{dy} = \frac{	an ^{-1} y - x}{1 + y^2}$ $\frac{dx}{dy} = \frac{	an ^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$ $\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{	an ^{-1} y}{1 + y^2}$ Here,$P(y) = \frac{1}{1 + y^2}$. 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 $A$.

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

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