An integrating factor of the differential equ | vedclass

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Question

(C) The given differential equation is $(1 - x^2)\frac{dy}{dx} - xy = 1$.Dividing both sides by $(1 - x^2)$,we get $\frac{dy}{dx} - \frac{x}{1 - x^2}y

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$\sqrt{1 - x^2}$

Vedclass · Answer

(C) The given differential equation is $(1 - x^2)\frac{dy}{dx} - xy = 1$.Dividing both sides by $(1 - x^2)$,we get $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{x}{1 - x^2}$ and $Q = \frac{1}{1 - x^2}$.The integrating factor $(I.F.)$ is given by $e^{\int P dx}$.$I.F. = e^{\int -\frac{x}{1 - x^2} dx}$.Let $u = 1 - x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.$I.F. = e^{\int \frac{1}{2u} du} = e^{\frac{1}{2}\log|u|} = e^{\log|u|^{1/2}} = \sqrt{1 - x^2}$.

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

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