The integrating factor of the differential eq | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{\sqrt{1-x^2}}$

Vedclass · Answer

(A) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = kx$. Dividing both sides by $(1-x^2)$,we get: $\frac{dy}{dx} + \frac{x}{1-x^2} y = \frac{kx}{1-x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{1-x^2}$. The integrating factor $(IF)$ is given by $e^{\int P(x) dx}$. $IF = e^{\int \frac{x}{1-x^2} dx}$. Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$. $IF = e^{-\frac{1}{2} \int \frac{1}{u} du} = e^{-\frac{1}{2} \ln|u|} = e^{\ln|u|^{-1/2}} = |u|^{-1/2} = \frac{1}{\sqrt{1-x^2}}$. Thus,the correct option is $A$.

The integrating factor of the differential equation $(1-x^2) \frac{dy}{dx} + xy = kx$ for $(-1 < x < 1)$ is . . . . . . .

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