An integrating factor of the differential equ | vedclass

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(D) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$. Dividing by $(1-x^2)$,we get: $\frac{dy}{d

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

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(D) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$. Dividing by $(1-x^2)$,we get: $\frac{dy}{dx} + \frac{x}{1-x^2} y = \frac{x^4 (\sqrt{1-x^2})^3}{(1+x^5)(1-x^2)} = \frac{x^4 \sqrt{1-x^2}}{1+x^5}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{x}{1-x^2}$. The integrating factor $(IF)$ is given by $e^{\int P dx}$. $IF = e^{\int \frac{x}{1-x^2} dx}$. Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$. $IF = e^{-\frac{1}{2} \int \frac{1}{u} du} = e^{-\frac{1}{2} \ln|u|} = e^{\ln|u|^{-1/2}} = u^{-1/2} = \frac{1}{\sqrt{1-x^2}}$.

An integrating factor of the differential equation $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$ is

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