Question

Integrating factor of the differential equation $(1 - x^2)\frac{dy}{dx} - xy = 1$ is

• A. $1 - x^2$
• B. $\frac{1}{2} \log (1 - x^2)$
• C. $\frac{x}{1 - x^2}$
• D. $\sqrt{1 - x^2}$

Hint:

Always remember to bring the differential equation to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ before identifying $P(x)$. Forgetting to divide by the coefficient of $\frac{dy}{dx}$ is a very common mistake.

Answer 1

The Correct Option is D

To find the integrating factor of the given differential equation, we start by rewriting the equation:

\((1 - x^2)\frac{dy}{dx} - xy = 1\).

First, we write it in the standard linear form:

\(\frac{dy}{dx} - \frac{xy}{1-x^2} = \frac{1}{1-x^2}\).

The standard linear form of a first-order differential equation is:

\(\frac{dy}{dx} + P(x)y = Q(x)\),

where \(P(x) = -\frac{x}{1-x^2}\) and \(Q(x) = \frac{1}{1-x^2}\).

The integrating factor (IF) is given by:

\(e^{\int P(x) \, dx}\)

We calculate \(\int P(x) \, dx\) as follows:

\(\int -\frac{x}{1-x^2} \, dx\)

We use substitution here. Let \(u = 1-x^2\), then \(\frac{du}{dx} = -2x\) or -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln|u| + C = -\frac{1}{2} \ln|1-x^2| + C

Thus, the integrating factor is:

\(e^{-\frac{1}{2} \ln|1-x^2|} = (1-x^2)^{-\frac{1}{2}}\)

In the context of the given options, this simplifies to:

\(\sqrt{1-x^2}\)

Thus, the integrating factor of the given differential equation is \(\sqrt{1-x^2}\). This matches the correct answer given in the options.