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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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$\left(1-x^2\right) \frac{d y}{d x}+x y=\frac{x^4}{\left(1+x^5\right)}\left(\sqrt{1-x^2}\right)^3$ is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{1-x^2}}$
Given differential equation can be rewritten as
$\begin{aligned} & \frac{d y}{d x}+\frac{x y}{1-x^2}=\frac{x^4\left(\sqrt{1-x^2}\right)^3}{\left(1+x^5\right)\left(1-x^2\right)} \\ & \frac{d y}{d x}+\frac{x y}{1-x^2}=\frac{x^4 \sqrt{1-x^2}}{\left(1+x^5\right)}\end{aligned}$
On comparing by linear differential equation is
$\begin{aligned} \frac{d y}{d x}+P y & =Q \\ \therefore \quad P & =\frac{x}{1-x^2}\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{IF}=e^{\int P d x} & =e^{\int \frac{x}{1-x^2} d x} \\ & =e^{-\frac{1}{2} \int \frac{-2 x}{1-x^2} d x}=e^{-\frac{1}{2} \log \left(1-x^2\right)} \\ & =e^{\log \left(1-x^2\right)^{-1 / 2}}=\frac{1}{\sqrt{1-x^2}}\end{aligned}$
$\begin{aligned} & \frac{d y}{d x}+\frac{x y}{1-x^2}=\frac{x^4\left(\sqrt{1-x^2}\right)^3}{\left(1+x^5\right)\left(1-x^2\right)} \\ & \frac{d y}{d x}+\frac{x y}{1-x^2}=\frac{x^4 \sqrt{1-x^2}}{\left(1+x^5\right)}\end{aligned}$
On comparing by linear differential equation is
$\begin{aligned} \frac{d y}{d x}+P y & =Q \\ \therefore \quad P & =\frac{x}{1-x^2}\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{IF}=e^{\int P d x} & =e^{\int \frac{x}{1-x^2} d x} \\ & =e^{-\frac{1}{2} \int \frac{-2 x}{1-x^2} d x}=e^{-\frac{1}{2} \log \left(1-x^2\right)} \\ & =e^{\log \left(1-x^2\right)^{-1 / 2}}=\frac{1}{\sqrt{1-x^2}}\end{aligned}$
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