The solution of `x log x (dy)/(dx) + y = 2/xlog x` is

Options

A. ` y log x = 1/x (1+logx)+C`
B. ` y log x = 2/x (1+log x)+C`
C. ` y log x = 2x (1+log x)+C`
D. ` y log x = (-2)/x (1+log x) +C`

Explanation

To solve the differential equation \( x \log x \frac{dy}{dx} + y = \frac{2}{x \log x} \), we will follow the steps outlined below:

### Step 1: Rewrite the Equation
First, we will rewrite the equation in a more manageable form by dividing through by \( x \log x \):
\[ \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^2 \log x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
In the standard form of a linear first-order differential equation \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:
- \( p(x) = \frac{1}{x \log x} \)
- \( q(x) = \frac{2}{x^2 \log x} \)

### Step 3: Find the Integrating Factor
Next, we find the integrating factor \( \mu(x) \) using the formula:
\[ \mu(x) = e^{\int p(x) \, dx} \]
Calculating the integral:
\[ \int \frac{1}{x \log x} \, dx \]
Let \( \log x = t \), then \( \frac{1}{x} dx = dt \) or \( dx = x dt = e^t dt \). Thus, we have:
\[ \int \frac{1}{x \log x} \, dx = \int \frac{1}{t} dt = \log |t| + C = \log |\log x| + C \]
So, the integrating factor becomes:
\[ \mu(x) = e^{\log |\log x|} = \log x \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by \( \log x \):
\[ \log x \frac{dy}{dx} + \frac{y \log x}{x \log x} = \frac{2 \log x}{x^2 \log x} \]
This simplifies to:
\[ \log x \frac{dy}{dx} + \frac{y}{x} = \frac{2}{x^2} \]

### Step 5: Rewrite the Left Side
The left side can be rewritten as:
\[ \frac{d}{dx}(y \log x) = \frac{2}{x^2} \]

### Step 6: Integrate Both Sides
Integrating both sides gives:
\[ y \log x = \int \frac{2}{x^2} \, dx \]
The integral evaluates to:
\[ \int \frac{2}{x^2} \, dx = -\frac{2}{x} + C \]
Thus, we have:
\[ y \log x = -\frac{2}{x} + C \]

### Step 7: Solve for \( y \)
Finally, we solve for \( y \):
\[ y = \frac{-\frac{2}{x} + C}{\log x} \]
This can be simplified to:
\[ y = \frac{-2}{x \log x} + \frac{C}{\log x} \]

### Final Answer
The solution to the differential equation is:
\[ y = \frac{-2}{x \log x} + \frac{C}{\log x} \]

---