The integrating factor of the differential eq | vedclass

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Question

(D) The given differential equation is $(1 - x^2) \frac{dy}{dx} - xy = 1$.Divide by $(1 - x^2)$ to get the standard linear form $\frac{dy}{dx} + P(x)y

Vedclass · Accepted Answer

$\sqrt{1 - x^2}$

Vedclass · Answer

(D) The given differential equation is $(1 - x^2) \frac{dy}{dx} - xy = 1$.Divide by $(1 - x^2)$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:$\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.Here, $P(x) = -\frac{x}{1 - x^2}$.The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{x}{1 - x^2} dx}$.Let $t = 1 - x^2$, then $dt = -2x dx$, which implies $-x dx = \frac{dt}{2}$.$I.F. = e^{\int \frac{dt}{2t}} = e^{\frac{1}{2} \log |t|} = e^{\log |t|^{1/2}} = \sqrt{t} = \sqrt{1 - x^2}$.

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

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