The integrating factor of differential euation `(1-x^(2))(dy)/(dx)-xy=1 " is"`

Options

A. `-x`
B. `(x)/(1+x^(2))`
C. `sqrt(1-x^(2))`
D. `(1)/(x)log(1-x^(2))`

Explanation

To find the integrating factor of the differential equation \((1 - x^2) \frac{dy}{dx} - xy = 1\), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ (1 - x^2) \frac{dy}{dx} - xy = 1 \]
To make it easier to work with, we can divide the entire equation by \(1 - x^2\):
\[ \frac{dy}{dx} - \frac{xy}{1 - x^2} = \frac{1}{1 - x^2} \]

### Step 2: Identify \(p\) and \(q\)
Now, we can identify \(p\) and \(q\) from the standard form of a linear differential equation:
\[ \frac{dy}{dx} + py = q \]
Here, we have:
\[ p = -\frac{x}{1 - x^2}, \quad q = \frac{1}{1 - x^2} \]

### Step 3: Calculate the Integrating Factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int p \, dx} = e^{\int -\frac{x}{1 - x^2} \, dx} \]
To solve the integral, we can use the substitution:
\[ t = 1 - x^2 \quad \Rightarrow \quad dt = -2x \, dx \quad \Rightarrow \quad -x \, dx = \frac{dt}{2} \]
Substituting this into the integral, we get:
\[ \int -\frac{x}{1 - x^2} \, dx = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{dt}{t} \]
The integral of \(\frac{1}{t}\) is:
\[ \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln |1 - x^2| + C \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{\frac{1}{2} \ln |1 - x^2|} = |1 - x^2|^{\frac{1}{2}} = \sqrt{1 - x^2} \]

### Final Result
The integrating factor of the differential equation \((1 - x^2) \frac{dy}{dx} - xy = 1\) is:
\[ \sqrt{1 - x^2} \]