Solution: The given differential equation is \[ x^2 y^{\prime\prime} - 4x y^\prime +6y =0, \] is Cauchy-Euler equation. The general form of Cauchy-Euler equation is \[ ax^2 y^{\prime\prime} + bxy^\prime + cy = 0. \] The solution is obtained by substituting $y=x^r$. The characteristic equation will be \[ ar(r-1) + br+c=0. \]

In the given problem, $a=1, b= -4, c = 6$. Thus, the characteristic equation will be \[ r(r-1)-4r+6 = 0 \implies r^2 -5r + 6 = 0. \] The roots of the above quadratic equation is $r=2,3$. Thus, the solution will be \[ y(x) = c_1x^2 + c_2 x^3. \] Now we will use the initial conditions to find the values of $c_1$ and $c_2$. \begin{align*} y(2) = 0 & \implies 4c_{1} + 8c_2 = 0 \implies c_1 + 2c_2 = 0 \\ y^\prime (2) = 4 & \implies 4c_1 + 12c_2 =4 \implies c_1 + 3c_2 = 1. \end{align*}

Solving the above system will give \[ c_1 = -2, \text{ and } c_2 = 1. \] Thus, the solution will be \[ y(x) = -2x^2 + x^3. \] Therefore, \[ y(4) = -32 + 64 = 32. \]