An integrating factor of the differential equ | vedclass

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(D) The given differential equation is $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$. Dividing bo

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

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(D) The given differential equation is $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$. Dividing both sides by $(1-x^2)$,we get: $\frac{d y}{d x} + \frac{x}{1-x^2} y = \frac{x^4 (1-x^2)^{3/2}}{(1+x^5)(1-x^2)} = \frac{x^4 \sqrt{1-x^2}}{1+x^5}$. This is a linear differential equation of the form $\frac{d y}{d x} + P(x) y = Q(x)$,where $P(x) = \frac{x}{1-x^2}$. The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int \frac{x}{1-x^2} dx}$. 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 $\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$ is

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