Write integrating factor differential equations
`x(dy)/(dx)+y log x= x+y`

Answer

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

### Step 1: Rewrite the equation in standard form
First, we need to rearrange the given equation into the standard linear form, which is:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
We start with:
\[ x \frac{dy}{dx} + y \log x = x + y \]
We can divide the entire equation by \( x \) to simplify it:
\[ \frac{dy}{dx} + \frac{y \log x}{x} = 1 + \frac{y}{x} \]
Rearranging gives us:
\[ \frac{dy}{dx} + \left(\frac{\log x - 1}{x}\right)y = 1 \]

### Step 2: Identify \( P(x) \) and \( Q(x) \)
From the standard form, we can identify:
- \( P(x) = \frac{\log x - 1}{x} \)
- \( Q(x) = 1 \)

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

### Step 4: Evaluate the integral
Now, we need to evaluate the integral:
\[ \int \left(\frac{\log x - 1}{x}\right) \, dx \]
We can split this into two parts:
\[ \int \frac{\log x}{x} \, dx - \int \frac{1}{x} \, dx \]
The first integral can be solved using integration by parts. Let \( u = \log x \) and \( dv = \frac{1}{x} \, dx \):
- Then, \( du = \frac{1}{x} \, dx \) and \( v = \log x \).

Using integration by parts:
\[ \int \log x \, d(\log x) = \log x \cdot \log x - \int \log x \cdot \frac{1}{x} \, dx \]
This results in:
\[ \frac{(\log x)^2}{2} - \log x \]
Thus:
\[ \int \frac{\log x}{x} \, dx = \frac{(\log x)^2}{2} \]
And the second integral is:
\[ \int \frac{1}{x} \, dx = \log x \]
Combining these results:
\[ \int \left(\frac{\log x - 1}{x}\right) \, dx = \frac{(\log x)^2}{2} - \log x \]

### Step 5: Substitute back to find the integrating factor
Now substituting back into the expression for the integrating factor:
\[ \mu(x) = e^{\frac{(\log x)^2}{2} - \log x} = e^{\frac{(\log x)^2}{2}} \cdot e^{-\log x} = \frac{e^{\frac{(\log x)^2}{2}}}{x} \]

### Final Answer
The integrating factor for the differential equation is:
\[ \mu(x) = \frac{e^{\frac{(\log x)^2}{2}}}{x} \]