The integrating factor of differential equation `(dy)/(dx)+y=(1+y)/(x)"is"`

Options

A. `(x)/(x^(x)`
B. `(e^(x))/(x)`
C. `xe^(x)`
D. `e^(x)`

Explanation

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

### Step 1: Rewrite the Differential Equation
We start by rewriting the given equation in the standard linear form \(\frac{dy}{dx} + py = q\).

\[ \frac{dy}{dx} + y = \frac{1 + y}{x} \]

Rearranging gives:

\[ \frac{dy}{dx} = \frac{1 + y}{x} - y \]

### Step 2: Simplify the Right Side
Now, we simplify the right-hand side:

\[ \frac{dy}{dx} = \frac{1 + y - yx}{x} \]

Taking the common denominator \(x\):

\[ \frac{dy}{dx} = \frac{1 - yx + y}{x} = \frac{1}{x} + \left( \frac{1 - x}{x} \right) y \]

### Step 3: Identify \(p\) and \(q\)
From the equation, we can identify:

\[ p = \frac{1 - x}{x}, \quad q = \frac{1}{x} \]

### Step 4: Find the Integrating Factor
The integrating factor \(\mu(x)\) is given by:

\[ \mu(x) = e^{\int p \, dx} \]

Substituting for \(p\):

\[ \mu(x) = e^{\int \frac{1 - x}{x} \, dx} \]

This can be split into two integrals:

\[ \mu(x) = e^{\int \left( \frac{1}{x} - 1 \right) \, dx} = e^{\int \frac{1}{x} \, dx - \int 1 \, dx} \]

### Step 5: Evaluate the Integrals
Calculating the integrals:

\[ \int \frac{1}{x} \, dx = \ln |x|, \quad \int 1 \, dx = x \]

Thus, we have:

\[ \mu(x) = e^{\ln |x| - x} = e^{\ln |x|} \cdot e^{-x} = |x| \cdot e^{-x} \]

### Step 6: Final Form of the Integrating Factor
Since we are typically interested in positive \(x\) for the context of differential equations, we can simplify to:

\[ \mu(x) = \frac{e^x}{x} \]

### Conclusion
The integrating factor of the differential equation is:

\[ \mu(x) = \frac{e^x}{x} \]