The integrating factor of the differential eq | vedclass

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Question

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

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$x^2 + 1$

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

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

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