Let `y = y (x)` be the solution of the differential equation `xlog_e x dy/dx+ y =x^2 log_ex(x>1)`. if y(2)=2, then y (e) is equal to

Options

A. `(4+e^2)/4`
B. `(1+e^2)/2`
C. `(2+e^2)/2`
D. `(1+e^2)/4`

Answer

To solve the given differential equation \( x \ln x \frac{dy}{dx} + y = x^2 \ln x \) with the initial condition \( y(2) = 2 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the differential equation in standard form:
\[ \frac{dy}{dx} + \frac{y}{x \ln x} = \frac{x^2 \ln x}{x \ln x} \]
This simplifies to:
\[ \frac{dy}{dx} + \frac{y}{x \ln x} = x \]

### Step 2: Identify the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int \frac{1}{x \ln x} \, dx} \]
To compute this integral, we can use the substitution \( t = \ln x \), which gives \( dt = \frac{1}{x} dx \). Thus, the integral becomes:
\[ \int \frac{1}{x \ln x} \, dx = \int \frac{1}{t} \, dt = \ln |t| + C = \ln |\ln x| + C \]
Therefore, the integrating factor is:
\[ \mu(x) = e^{\ln |\ln x|} = \ln x \]

### Step 3: Multiply the Equation by the Integrating Factor
Now we multiply the entire differential equation by \( \ln x \):
\[ \ln x \frac{dy}{dx} + \frac{y \ln x}{x \ln x} = x \ln x \]
This simplifies to:
\[ \ln x \frac{dy}{dx} + \frac{y}{x} = x \ln x \]

### Step 4: Recognize the Left Side as a Derivative
The left-hand side can be recognized as the derivative of a product:
\[ \frac{d}{dx}(y \ln x) = x \ln x \]

### Step 5: Integrate Both Sides
Integrating both sides gives:
\[ y \ln x = \int x \ln x \, dx \]
Using integration by parts, let \( u = \ln x \) and \( dv = x \, dx \):
\[ du = \frac{1}{x} \, dx, \quad v = \frac{x^2}{2} \]
Thus,
\[ \int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \]
So we have:
\[ y \ln x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \]

### Step 6: Solve for \( y \)
Dividing by \( \ln x \):
\[ y = \frac{x^2}{2} - \frac{x^2}{4 \ln x} + \frac{C}{\ln x} \]

### Step 7: Use the Initial Condition
Using the condition \( y(2) = 2 \):
\[ 2 = \frac{2^2}{2} - \frac{2^2}{4 \ln 2} + \frac{C}{\ln 2} \]
This simplifies to:
\[ 2 = 2 - \frac{4}{4 \ln 2} + \frac{C}{\ln 2} \]
\[ 0 = -1 + \frac{C}{\ln 2} \]
Thus, \( C = \ln 2 \).

### Step 8: Substitute \( C \) Back into the Equation
Now substituting \( C \) back:
\[ y = \frac{x^2}{2} - \frac{x^2}{4 \ln x} + \frac{\ln 2}{\ln x} \]

### Step 9: Find \( y(e) \)
Now we need to find \( y(e) \):
\[ y(e) = \frac{e^2}{2} - \frac{e^2}{4 \cdot 1} + \frac{\ln 2}{1} \]
This simplifies to:
\[ y(e) = \frac{e^2}{2} - \frac{e^2}{4} + \ln 2 = \frac{2e^2}{4} - \frac{e^2}{4} + \ln 2 = \frac{e^2}{4} + \ln 2 \]

### Final Answer
Thus, the value of \( y(e) \) is:
\[ \boxed{\frac{e^2}{4} + \ln 2} \]