The integrating factor of differential equation (1 - x^2) dy/dx - xy…

The correct answer is Option (3) → $\sqrt{1 - x^2}$ ##

Given that, $(1 - x^2) \frac{dy}{dx} - xy = 1$

$\Rightarrow \frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{1}{1 - x^2}$

which is a linear differential equation and on comparing with $\frac{dy}{dx} + P \cdot y = Q$, we get

$P = \frac{-x}{1 - x^2}$

$Q = \frac{1}{1 - x^2}$

$∴\text{I.F} = e^{\int \frac{-x}{1 - x^2} \, dx} \quad [\text{I.F} = e^{\int P \, dx}]$

Put $1 - x^2 = t \Rightarrow -2x \, dx = dt \Rightarrow x \, dx = -\frac{dt}{2}$

$\text{I.F} = e^{\frac{1}{2} \int \frac{dt}{t}} = e^{\frac{1}{2} \log t}$

Now, $\text{I.F} = e^{\frac{1}{2} \log(1 - x^2)} = \sqrt{1 - x^2} \quad [∵1 - x^2 = t]$