The given differential equation is:

\[
y'' - 4y' + 4y = \sin(x) e^{2x}
\]

This is a second-order non-homogeneous linear differential equation. The solution will consist of two parts:

1. **The general solution of the homogeneous equation** \( y'' - 4y' + 4y = 0 \).
2. **A particular solution** to the non-homogeneous equation \( y'' - 4y' + 4y = \sin(x) e^{2x} \).

### Step 1: Solve the homogeneous equation
The homogeneous equation is:

\[
y'' - 4y' + 4y = 0
\]

This is a linear differential equation with constant coefficients. The characteristic equation is:

\[
r^2 - 4r + 4 = 0
\]

Solving this quadratic equation:

\[
r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)}
\]
\[
r = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2
\]

So, \( r = 2 \) is a repeated root. Thus, the general solution to the homogeneous equation is:

\[
y_h(x) = (C_1 + C_2 x) e^{2x}
\]

### Step 2: Find the particular solution
To find the particular solution \( y_p(x) \), we use the method of undetermined coefficients. The non-homogeneous term is \( \sin(x) e^{2x} \). Given that the homogeneous solution contains terms involving \( e^{2x} \), we modify our guess for the particular solution.

We try the following form for the particular solution:

\[
y_p(x) = e^{2x} (A \sin(x) + B \cos(x))
\]

We will differentiate this expression to substitute into the original equation and solve for \( A \) and \( B \).

Let's compute the derivatives \( y_p' \) and \( y_p'' \), and then substitute them into the differential equation.from sympy import symbols, sin, cos, exp, diff, Eq, solve

# Define the symbols
x = symbols('x')
A, B = symbols('A B')

# Define the particular solution
y_p = exp(2*x) * (A*sin(x) + B*cos(x))

# Calculate first and second derivatives
y_p_prime = diff(y_p, x)
y_p_double_prime = diff(y_p_prime, x)

# Define the original equation y'' - 4y' + 4y = sin(x)e^(2x)
lhs = y_p_double_prime - 4*y_p_prime + 4*y_p
rhs = sin(x) * exp(2*x)

# Solve for A and B by equating coefficients
equation = Eq(lhs, rhs)
equation_simplified = equation.simplify()
equation_simplifiedThe simplified equation obtained after substituting the particular solution and its derivatives into the original differential equation is:

\[
e^{2x} \sin(x) = -e^{2x} (A \sin(x) + B \cos(x))
\]

We can factor out \( e^{2x} \) on both sides:

\[
\sin(x) = -A \sin(x) - B \cos(x)
\]

To satisfy this equation for all \( x \), we compare the coefficients of \( \sin(x) \) and \( \cos(x) \) on both sides:

- Coefficient of \( \sin(x) \): \( 1 = -A \)
- Coefficient of \( \cos(x) \): \( 0 = -B \)

From this, we get:
\[
A = -1, \quad B = 0
\]

Thus, the particular solution is:

\[
y_p(x) = -e^{2x} \sin(x)
\]

### Final Solution
The general solution to the given differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = (C_1 + C_2 x) e^{2x} - e^{2x} \sin(x)
\]

Would you like any details on the steps or to see more examples? Here are a few related questions to expand on:

1. How do we determine the form of the particular solution?
2. What happens if the non-homogeneous term is different, like \( e^x \)?
3. Why did we have to modify the particular solution form due to the homogeneous solution?
4. Can undetermined coefficients be applied to all non-homogeneous terms?
5. How do we solve higher-order linear differential equations?
6. What is the importance of the characteristic equation in solving differential equations?
7. Can the method of variation of parameters be used here?
8. How do the initial conditions affect the final solution?

**Tip:** Always check if the non-homogeneous term is similar to the solution of the homogeneous equation, as this affects the form of the particular solution.

Learn how to solve the differential equation y'' - 4y' + 4y = sin(x)e^2x with detailed steps. Understand the method of undetermined coefficients and the general solution of second-order linear differential equations. Suitable for advanced undergraduate students and anyone studying differential equations.

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