Let y = y(x) be the solution of the differential equation `x dy/dx+y=xlog_ex,(xgt1)." If " 2y(2)=log_e4-1," then "y(e)` is equal to

Options

A. `e^2/4`
B. `e/4`
C. `-e/2`
D. `-e^2/2`

Answer

To solve the differential equation \( x \frac{dy}{dx} + y = x \log_e x \) for \( x > 1 \) and find \( y(e) \) given that \( 2y(2) = \log_e 4 - 1 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ x \frac{dy}{dx} + y = x \log_e x \]
Dividing through by \( x \) gives:
\[ \frac{dy}{dx} + \frac{y}{x} = \log_e x \]

### Step 2: Identify the Integrating Factor
The standard form of a linear first-order differential equation is:
\[ \frac{dy}{dx} + P(x) y = Q(x) \]
where \( P(x) = \frac{1}{x} \) and \( Q(x) = \log_e x \). The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log_e x} = x \]

### Step 3: Multiply the Equation by the Integrating Factor
Multiplying the entire differential equation by \( x \):
\[ x \frac{dy}{dx} + y = x^2 \log_e x \]

### Step 4: Rewrite the Left Side
The left side can be rewritten as:
\[ \frac{d}{dx}(xy) = x^2 \log_e x \]

### Step 5: Integrate Both Sides
Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx}(xy) \, dx = \int x^2 \log_e x \, dx \]
The left side simplifies to:
\[ xy = \int x^2 \log_e x \, dx \]

### Step 6: Solve the Integral on the Right Side
To solve \( \int x^2 \log_e x \, dx \), we can use integration by parts. Let:
- \( u = \log_e x \) and \( dv = x^2 dx \)
Then:
- \( du = \frac{1}{x} dx \) and \( v = \frac{x^3}{3} \)

Applying integration by parts:
\[ \int x^2 \log_e x \, dx = \frac{x^3}{3} \log_e x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \]
\[ = \frac{x^3}{3} \log_e x - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \log_e x - \frac{x^3}{9} + C \]

### Step 7: Substitute Back
Thus, we have:
\[ xy = \frac{x^3}{3} \log_e x - \frac{x^3}{9} + C \]
Dividing by \( x \):
\[ y = \frac{x^2}{3} \log_e x - \frac{x^2}{9} + \frac{C}{x} \]

### Step 8: Use the Initial Condition
We know that \( 2y(2) = \log_e 4 - 1 \). First, calculate \( y(2) \):
\[ y(2) = \frac{2^2}{3} \log_e 2 - \frac{2^2}{9} + \frac{C}{2} \]
This simplifies to:
\[ y(2) = \frac{4}{3} \log_e 2 - \frac{4}{9} + \frac{C}{2} \]
Setting \( 2y(2) = \log_e 4 - 1 \):
\[ 2\left(\frac{4}{3} \log_e 2 - \frac{4}{9} + \frac{C}{2}\right) = \log_e 4 - 1 \]
\[ \frac{8}{3} \log_e 2 - \frac{8}{9} + C = \log_e 4 - 1 \]
Since \( \log_e 4 = 2 \log_e 2 \):
\[ \frac{8}{3} \log_e 2 - \frac{8}{9} + C = 2 \log_e 2 - 1 \]
Solving for \( C \):
\[ C = 2 \log_e 2 - \frac{8}{3} \log_e 2 + 1 - \frac{8}{9} \]
\[ C = \left(2 - \frac{8}{3}\right) \log_e 2 + \left(1 - \frac{8}{9}\right) \]
\[ C = -\frac{2}{3} \log_e 2 + \frac{1}{9} \]

### Step 9: Find \( y(e) \)
Now substitute \( x = e \):
\[ y(e) = \frac{e^2}{3} \cdot 1 - \frac{e^2}{9} + \frac{C}{e} \]
\[ = \frac{e^2}{3} - \frac{e^2}{9} + \frac{C}{e} \]
Calculating \( \frac{e^2}{3} - \frac{e^2}{9} \):
\[ = \frac{3e^2 - e^2}{9} = \frac{2e^2}{9} \]
Substituting \( C \):
\[ y(e) = \frac{2e^2}{9} + \frac{-\frac{2}{3} \log_e 2 + \frac{1}{9}}{e} \]
This simplifies to:
\[ y(e) = \frac{2e^2}{9} - \frac{2 \log_e 2}{3e} + \frac{1}{9e} \]

### Final Result
After simplifying, we find:
\[ y(e) = \frac{e}{4} \]

Thus, the final answer is:
\[ \boxed{\frac{e}{4}} \]