Integrating factor of the equation `(x^2+1)"dy"/"dx"+2xy=x^2-1` is

Options

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

Answer

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

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ (x^2 - 1) \frac{dy}{dx} + 2xy = x^2 - 1 \]
To put it in standard linear form, we divide the entire equation by \(x^2 - 1\) (assuming \(x^2 - 1 \neq 0\)):
\[ \frac{dy}{dx} + \frac{2xy}{x^2 - 1} = 1 \]

### Step 2: Identify \(p(x)\) and \(q(x)\)
From the standard form \(\frac{dy}{dx} + p(x)y = q(x)\), we can identify:
- \(p(x) = \frac{2x}{x^2 - 1}\)
- \(q(x) = 1\)

### Step 3: Calculate the integrating factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \frac{2x}{x^2 - 1} \, dx} \]

### Step 4: Simplify the integral
To solve the integral \(\int \frac{2x}{x^2 - 1} \, dx\), we can use substitution. Let:
\[ t = x^2 - 1 \quad \Rightarrow \quad dt = 2x \, dx \]
Thus, the integral becomes:
\[ \int \frac{2x}{x^2 - 1} \, dx = \int \frac{dt}{t} = \ln |t| + C = \ln |x^2 - 1| + C \]

### Step 5: Find the integrating factor
Now substituting back, we have:
\[ \mu(x) = e^{\ln |x^2 - 1|} = |x^2 - 1| \]
Since we are typically interested in the positive integrating factor, we can write:
\[ \mu(x) = x^2 - 1 \]

### Final Result
Thus, the integrating factor of the given differential equation is:
\[ \mu(x) = x^2 - 1 \]

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