Let ϕ denote the solution to the boundary value problem (BVP) 

\(\rm \left\{\begin{matrix}(xy')'-2y'+\frac{y}{x}=1,&1<x<e^4\\\ y(1)=0, y(e^4)=4e^4\end{matrix}\right.\)

Then the value of ϕ(e) is

Concept: 

Second-Order Differential Equations: Solve the homogeneous equation first, then find the particular solution.

Boundary Value Problems: Use boundary conditions to determine the constants in the general solution.

Explanation:
\((xy')' - 2y' + \frac{y}{x} = 1, \quad 1 < x < e^4\)

with boundary conditions \(y(1) = 0, \quad y(e^4) = 4e^4\)

Rewriting the equation:

The equation is simplified to

 \(x^2 y'' - xy' + y = x\)

Let x = e^z then it becomes

⇒ {D'(D' - 1) - D' + 1}y = e^z

⇒ (D'² - 2D' + 1)y = e^z

Auxiliary equation is

m² - 2m + 1 = 0

⇒ \((m-1)^2 = 0\)

⇒ \(m = 1,1 \)  

CF = C₁e^z + C₂ze^z 

 i.e., CF = \( C_1 x + C_2 x \ln(x)\)

PI = \({1\over (D'^2-2D'+1)}e^z\)

   = \(e^z{1\over ((D'+1)^2-2(D'+1)+1)}.1\)

  = \(e^z{1\over D'^2}.1\)

  = \({z^2\over 2}e^z\) = \(\frac {x(\ln x)^2}{2}\)
 
\( y(x) = C_1 x + C_2 x \ln(x) \)+ \(\frac {x(ln x)^2}{2}\)

Applying Boundary Conditions:

y(1) = 0:

 \(C_1 \cdot 1 + C_2 \cdot 1 \ln(1) + \frac{1^2}{2}.0 = 0\)
 
⇒ \(C_1 = 0\)

\(y(e^4) = 4e^4 \): 

\(\frac{1}{2} e^4 + C_2 e^4 \ln(e^4) + e^4\frac{(ln e^4)^2}{2} = 4e^4\)
   
⇒  \( C_2 e^4 \cdot 4 + 8{e^4} = 4e^4\)

⇒ \(c_2 = -1\)
   
\(y(x) = -x ln x + \)\(\frac {x(ln x)^2}{2}\)

⇒ \(y(e) = -e +\frac{e}{2}\)

⇒ \(y(e) = - \frac{e}{2}\)

Hence option 1) is correct.