Q.Integrating factor of the differential equation $(1-x^2)\frac{dy}{dx}-xy=1$ is:
(A) $-x$
(B) $\frac{x}{1+x^2}$
(C) $\sqrt{1-x^2}$
(D) $\frac{1}{2}\log(1-x^2)$

The key idea is to rewrite the equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ and then compute the integrating factor $\mu = e^{\int P\,dx}$.

First, divide through by $1-x^2$:

Here $P(x) = -\frac{x}{1-x^2}$.

Now compute $\int P\,dx$:

$\int -\frac{x}{1-x^2}\,dx = \frac{1}{2}\int \frac{-2x}{1-x^2}\,dx = \frac{1}{2}\log|1-x^2|.$ …

The integrating factor method is the go-to tool for first-order linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The idea: multiply the whole equation by a cleverly chosen function $\mu(x)$ so that the left-hand side becomes the derivative of $\mu(x) y$. That turns the problem into a simple integration.

Here, the given equation is $(1-x^2)\frac{dy}{dx} - xy = 1$. It’s not yet in standard form — the coefficient of $\frac{dy}{dx}$ is $1-x^2$, not 1. So the first step is always to divide through by that coefficient.

1. Rewrite in standard linear form

Divide every term by $1-x^2$ (valid for $x \neq \pm 1$, which we assume for the domain):

Now it matches $\frac{dy}{dx} + P(x)y = Q(x)$ with

2. Recall the formula for the integrating factor

So we need $\int P(x)\,dx = \int \frac{-x}{1-x^2}\,dx$.

3. Compute the integral

Let $u = 1 - x^2$. Then $du = -2x\,dx$, so $x\,dx = -\frac{1}{2}du$. Substituting:

Back-substitute $u = 1 - x^2$:

We only need one antiderivative (the constant can be ignored for the integrating factor), so take $C = 0$.

4. Form the integrating factor

$\mu(x) = e^{\frac{1}{2} \log|1 - x^2|} = |1 - x^2|^{1/2}.$ …

Method: Integrating Factor Involving a $\frac{1}{2}\log$ Integral

Some integrating factors require integrating $P(x)$ into a half-logarithm, which then exponentiates to a square root. Handle the $\frac{1}{2}$ factor carefully.

Steps

Step 1: Normalise and read $P(x)$.

For $(1-x^2)\frac{dy}{dx}-xy=1$, divide by $1-x^2$: $P(x)=-\frac{x}{1-x^2}$.

Step 2: Integrate $P$, pulling out $\frac{1}{2}$. …

Common Mistakes

Mistake 1: Forgetting the $\frac{1}{2}$ when integrating $-\frac{x}{1-x^2}$.

Why it's wrong: the integral is $\frac{1}{2}\log|1-x^2|$ because $\frac{d}{dx}(1-x^2)=-2x$; dropping the $\frac{1}{2}$ gives $1-x^2$ instead of its square root. Correct approach: factor out $\frac{1}{2}$, so the integrating factor is $\sqrt{1-x^2}$.

Mistake 2: Leaving the integrating factor as a logarithm. …