The solution of the differential equations $\left(1-x^2\right)\frac{dy}{dx}-xy=1$ (where, $\left|x\right| <1, x\in R \mathrm{and} \mathrm{C}$ is an arbitrary constant )

Given the differential equation $\left(1-{\mathrm{x}}^2\right)\frac{d\mathrm{y}}{d\mathrm{x}}-\mathrm{xy}=1$.

Divide the given Equation by $\left(1-{\mathrm{x}}^2\right)$,

$\Rightarrow \frac{d\mathrm{y}}{d\mathrm{x}}-\frac{\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{y}=\frac{1}{1-{\mathrm{x}}^2}$

The resulting Equation will be a Linear Differential Equation of type $\frac{d\mathrm{y}}{d\mathrm{x}} + P\left(x\right)y = \mathrm{Q}\left(\mathrm{x}\right)$

And solution to a linear differential Equation is $\mathrm{y}\times \mathrm{IF} = \int \mathrm{Q}\times \mathrm{IF}\times \mathrm{dx} , \left[\mathrm{IF}= {\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\right]$

$$\begin{gathered} \mathrm{I}.\mathrm{F}.= {\mathrm{e}}^{-\int \frac{\mathrm{x}}{1-{\mathrm{x}}^2}d\mathrm{x}} \\ \Rightarrow \mathrm{IF} ={\mathrm{e}}^{\frac{1}{2}\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}d\mathrm{x} } \end{gathered}$$

For $\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}$ ,

Since , $\mathrm{d}\left(1-{\mathrm{x}}^2\right)=-2\mathrm{xdx}$

$\Rightarrow \int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx} = \int \frac{-\mathrm{d}\left(1-{\mathrm{x}}^2\right)}{1-{\mathrm{x}}^2}$

As , $\int \frac{\mathrm{dt}}{\mathrm{t}}=\ln \left(\mathrm{t}\right)$,

$\Rightarrow \int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx} = -\ln \left(1-{\mathrm{x}}^2\right)+\mathrm{c}$

Replace the value of $\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}$ in $\mathrm{IF}$.

$\Rightarrow \mathrm{IF}={\mathrm{e}}^{\frac{-1}{2}\mathrm{In}\left(1-{\mathrm{x}}^2\right)=-\sqrt{1-{\mathrm{x}}^2}}$

Thus , solution of the differential equation is,

$$\begin{gathered} -\mathrm{y}\sqrt{1-{\mathrm{x}}^2}=\int \frac{-\sqrt{1-{\mathrm{x}}^2}}{1-{\mathrm{x}}^2}d\mathrm{x} \\ \mathrm{y}\sqrt{1-{\mathrm{x}}^2}=\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}d\mathrm{x} \\ \mathrm{y}\sqrt{1-{\mathrm{x}}^2}={\sin }^{-1}\left(\mathrm{x}\right)+\mathrm{c} \end{gathered}$$

Hence option-3 is the correct answer.