The integrating factor of the differential eq | vedclass

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Question

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

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$x^2 - 1$

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(B) The given differential equation is $(x^2 - 1)\frac{dy}{dx} + 2xy = x$. Dividing both sides by $(x^2 - 1)$,we get $\frac{dy}{dx} + \frac{2x}{x^2 - 1}y = \frac{x}{x^2 - 1}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{x^2 - 1}$ and $Q = \frac{x}{x^2 - 1}$. The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{2x}{x^2 - 1} dx}$. Let $t = x^2 - 1$,then $dt = 2x dx$. Thus,$IF = e^{\int \frac{dt}{t}} = e^{\ln|t|} = t = x^2 - 1$. Therefore,the integrating factor is $x^2 - 1$.

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

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