The solution of the differential equation `ydx-xdy+lnxdx=0` is (where, C is an arbitrary constant)

Options

A. `y=(ln x)^(2)+C`
B. `y=(ln x+1)+C`
C. `y=-( ln x+1)+C`
D. `y=(lnx)(x+C)`

Answer

To solve the differential equation \( y \, dx - x \, dy + \ln x \, dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ y \, dx - x \, dy + \ln x \, dx = 0 \]
This can be rewritten as:
\[ (y + \ln x) \, dx = x \, dy \]

### Step 2: Dividing by \( dx \)
Next, we divide both sides by \( dx \):
\[ y + \ln x = x \frac{dy}{dx} \]
Rearranging gives us:
\[ x \frac{dy}{dx} - y = \ln x \]

### Step 3: Identifying the Form
We can identify this as a linear first-order differential equation of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = -\frac{1}{x} \) and \( Q(x) = \frac{\ln x}{x} \).

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is calculated as:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} \]

### Step 5: Multiplying the Equation by the Integrating Factor
We multiply the entire differential equation by the integrating factor:
\[ \frac{1}{x} \left( x \frac{dy}{dx} - y \right) = \frac{\ln x}{x^2} \]
This simplifies to:
\[ \frac{dy}{dx} - \frac{y}{x} = \frac{\ln x}{x^2} \]

### Step 6: Integrating Both Sides
Now we integrate both sides:
\[ \int \left( \frac{dy}{dx} - \frac{y}{x} \right) dx = \int \frac{\ln x}{x^2} \, dx \]
The left side integrates to:
\[ y = \int \frac{\ln x}{x^2} \, dx + C \]

### Step 7: Solving the Right Side
To solve the integral on the right side, we use integration by parts:
Let \( u = \ln x \) and \( dv = \frac{1}{x^2} dx \).
Then \( du = \frac{1}{x} dx \) and \( v = -\frac{1}{x} \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives us:
\[ \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \int -\frac{1}{x^2} \, dx = -\frac{\ln x}{x} + \frac{1}{x} + C \]

### Step 8: Final Solution
Substituting back, we have:
\[ y = -\frac{\ln x}{x} + \frac{1}{x} + C \]
Rearranging gives us:
\[ y = -\ln x + 1 + Cx \]

### Final Answer
Thus, the solution of the differential equation is:
\[ y = -\ln x + 1 + C \]