Find the integrating factor of the given differential equation: `ydx-xdy+(logx)dx=0`

Answer

To find the integrating factor of the given differential equation \( ydx - xdy + \log x \, dx = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ ydx - xdy + \log x \, dx = 0 \]
We can rearrange this equation as:
\[ ydx + \log x \, dx = xdy \]

### Step 2: Divide by \( dx \)
Next, we divide the entire equation by \( dx \):
\[ y + \log x = x \frac{dy}{dx} \]

### Step 3: Rearrange to standard form
Now, we can rearrange this equation into the standard form of a first-order linear differential equation:
\[ \frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x} \]

### Step 4: Identify \( p \)
In the standard form \( \frac{dy}{dx} + p y = q \), we identify \( p \):
\[ p = -\frac{1}{x} \]

### Step 5: Calculate the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p \, dx} = e^{\int -\frac{1}{x} \, dx} \]
Calculating the integral:
\[ \int -\frac{1}{x} \, dx = -\ln |x| = \ln |x|^{-1} \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{\ln |x|^{-1}} = |x|^{-1} = \frac{1}{x} \]

### Step 6: Final Answer
The integrating factor of the given differential equation is:
\[ \frac{1}{x} \]

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