Solution: The given differential equation is: \[ y'' - 4y' + 4y = e^{2x}. \] The homogeneous equation corresponding to the given ODE is: \[ y'' - 4y' + 4y = 0. \] The characteristic equation is: \[ r^2 - 4r + 4 = 0. \] Solving this, we get a repeated root: \[ r = 2 \quad \text{(with multiplicity 2)}. \] Therefore, the general solution to the homogeneous equation is: \[ y_h(x) = (C_1 + C_2 x) e^{2x}, \] where \( C_1 \) and \( C_2 \) are arbitrary constants.

Since the right-hand side of the equation is \( e^{2x} \) and \( r = 2 \) is a repeated root of the characteristic equation, we assume a particular solution of the form: \[ y_p(x) = A x^2 e^{2x}, \] where \( A \) is a constant to be determined. Differentiating \( y_p(x) \) twice, we get: \[ y_p'(x) = A (2x + 2x^2) e^{2x}, \] \[ y_p''(x) = A (2 + 8x + 4x^2) e^{2x}. \] Substituting \( y_p(x) \), \( y_p'(x) \), and \( y_p''(x) \) into the original equation, we obtain: \[ A (2 + 8x + 4x^2) e^{2x} - 4A (2x + 2x^2) e^{2x} + 4A x^2 e^{2x} = e^{2x}. \] Simplifying, we find: \[ 2A e^{2x} = e^{2x}. \] Thus, \( A = \frac{1}{2} \), and the particular solution is: \[ y_p(x) = \frac{1}{2} x^2 e^{2x}. \]

The general solution to the non-homogeneous equation is the sum of the homogeneous solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) = (C_1 + C_2 x) e^{2x} + \frac{1}{2} x^2 e^{2x}. \]