Let y(x) be the solution of the differential equation `(xlogx)(dy)/(dx)+y=2xlogx, (xge1)`, Then y(e) is equal to

Options

A. e
B. 0
C. 2
D. 2e

Answer

To solve the differential equation \((x \log x) \frac{dy}{dx} + y = 2x \log x\) for \(x \geq 1\) and find \(y(e)\), we will follow these steps:

### Step 1: Rewrite the differential equation
We start with the given equation:
\[ (x \log x) \frac{dy}{dx} + y = 2x \log x \]
To simplify, divide through by \(x \log x\):
\[ \frac{dy}{dx} + \frac{y}{x \log x} = 2 \]

### Step 2: Identify \(p\) and \(q\)
This equation is now in the standard form:
\[ \frac{dy}{dx} + p y = q \]
where \(p = \frac{1}{x \log x}\) and \(q = 2\).

### Step 3: Find the integrating factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int p \, dx} = e^{\int \frac{1}{x \log x} \, dx} \]
To solve this integral, we use the substitution \(t = \log x\), which gives \(dt = \frac{1}{x} dx\). Thus, the integral becomes:
\[ \int \frac{1}{x \log x} \, dx = \int \frac{1}{t} \, dt = \log t = \log(\log x) \]
Therefore, the integrating factor is:
\[ \mu(x) = e^{\log(\log x)} = \log x \]

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

### Step 5: Rewrite the left-hand side
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y \log x) = 2 \log x \]

### Step 6: Integrate both sides
Integrating both sides with respect to \(x\):
\[ \int \frac{d}{dx}(y \log x) \, dx = \int 2 \log x \, dx \]
The left-hand side gives:
\[ y \log x = \int 2 \log x \, dx \]
To solve the integral on the right, we use integration by parts:
Let \(u = \log x\) and \(dv = 2 \, dx\), then \(du = \frac{1}{x} dx\) and \(v = 2x\):
\[ \int 2 \log x \, dx = 2x \log x - \int 2x \cdot \frac{1}{x} \, dx = 2x \log x - 2x + C \]
Thus,
\[ y \log x = 2x \log x - 2x + C \]

### Step 7: Solve for \(y\)
Dividing by \(\log x\):
\[ y = 2x - \frac{2x}{\log x} + \frac{C}{\log x} \]

### Step 8: Find the constant \(C\)
To find \(C\), we can use the initial condition. We set \(x = 1\):
\[ y(1) \log 1 = 2(1) \log 1 - 2(1) + C \]
Since \(\log 1 = 0\), we have:
\[ 0 = 0 - 2 + C \implies C = 2 \]

### Step 9: Substitute \(C\) back into the equation
Now substituting \(C\) back:
\[ y = 2x - \frac{2x}{\log x} + \frac{2}{\log x} \]

### Step 10: Evaluate \(y(e)\)
Now we need to find \(y(e)\):
\[ y(e) = 2e - \frac{2e}{\log e} + \frac{2}{\log e} \]
Since \(\log e = 1\):
\[ y(e) = 2e - 2e + 2 = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]