An integrating factor of the differential equ | vedclass

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Question

(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 - x^2}$ and $Q = \frac{x}{\sqrt{1 - x^2}}$.The inte

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$(1 - x^2)^{-1}$

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(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 - x^2}$ and $Q = \frac{x}{\sqrt{1 - x^2}}$.The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx}$.$I.F. = e^{\int \frac{2x}{1 - x^2} dx}$.Let $u = 1 - x^2$,then $du = -2x dx$,so $2x dx = -du$.$I.F. = e^{\int \frac{-du}{u}} = e^{-\ln|u|} = e^{-\ln|1 - x^2|} = e^{\ln|(1 - x^2)^{-1}|}$.$I.F. = (1 - x^2)^{-1}$.

An integrating factor of the differential equation $\frac{dy}{dx} + \frac{2xy}{1 - x^2} = \frac{x}{\sqrt{1 - x^2}}$ is

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