To solve the differential equation \( y'' - 2y' + y = x + 1 + \sin(2x) \), we can use the method of undetermined coefficients. This is a second-order linear non-homogeneous differential equation. Here's the step-by-step solution:

### Step 1: Solve the Homogeneous Equation
First, solve the corresponding homogeneous equation:
\[
y'' - 2y' + y = 0
\]
The characteristic equation is:
\[
r^2 - 2r + 1 = 0
\]
This can be factored as:
\[
(r - 1)^2 = 0
\]
So, we have a repeated root \( r = 1 \). Therefore, the general solution to the homogeneous equation is:
\[
y_h(x) = (C_1 + C_2 x) e^x
\]
where \( C_1 \) and \( C_2 \) are constants.

### Step 2: Find a Particular Solution
The particular solution \( y_p(x) \) will be a combination of the solutions to the non-homogeneous terms \( x + 1 \) and \( \sin(2x) \).

For the \( x + 1 \) part, assume:
\[
y_p^{(1)}(x) = Ax + B
\]
For the \( \sin(2x) \) part, assume:
\[
y_p^{(2)}(x) = C\sin(2x) + D\cos(2x)
\]
So the total particular solution is:
\[
y_p(x) = Ax + B + C\sin(2x) + D\cos(2x)
\]

### Step 3: Substitute into the Original Equation
Substitute \( y_p(x) \) into the original differential equation:
\[
y''_p - 2y'_p + y_p = x + 1 + \sin(2x)
\]
Compute the derivatives:
- \( y_p'(x) = A + 2C\cos(2x) - 2D\sin(2x) \)
- \( y_p''(x) = -4C\sin(2x) - 4D\cos(2x) \)

Substitute these into the differential equation:
\[
(-4C\sin(2x) - 4D\cos(2x)) - 2(A + 2C\cos(2x) - 2D\sin(2x)) + (Ax + B + C\sin(2x) + D\cos(2x)) = x + 1 + \sin(2x)
\]

Simplify and group the coefficients:
- Coefficients of \( x \): \( A = 1 \)
- Constant terms: \( B - 2A = 1 \Rightarrow B = 3 \)
- Coefficients of \( \sin(2x) \): \( -4C + 4D + C = 1 \Rightarrow -3C + 4D = 1 \)
- Coefficients of \( \cos(2x) \): \( -4D - 4C + D = 0 \Rightarrow -4C - 3D = 0 \)

Solve the system of equations for \( C \) and \( D \):
- From \( -4C - 3D = 0 \), \( C = -\frac{3}{4}D \)
- Substitute into \( -3C + 4D = 1 \): \( \frac{9}{4}D + 4D = 1 \Rightarrow D = \frac{4}{25}, C = -\frac{3}{25} \)

Thus, the particular solution is:
\[
y_p(x) = x + 3 - \frac{3}{25}\sin(2x) + \frac{4}{25}\cos(2x)
\]

### Step 4: Write the General Solution
The general solution is the sum of the homogeneous and particular solutions:
\[
y(x) = y_h(x) + y_p(x) = (C_1 + C_2 x) e^x + x + 3 - \frac{3}{25}\sin(2x) + \frac{4}{25}\cos(2x)
\]

### Summary
The general solution to the differential equation \( y'' - 2y' + y = x + 1 + \sin(2x) \) is:
\[
y(x) = (C_1 + C_2 x) e^x + x + 3 - \frac{3}{25}\sin(2x) + \frac{4}{25}\cos(2x)
\]
where \( C_1 \) and \( C_2 \) are arbitrary constants.

Do you want more details or have any questions?

Here are some related questions:
1. How do you solve non-homogeneous differential equations using the variation of parameters method?
2. What is the difference between a homogeneous and non-homogeneous differential equation?
3. How do repeated roots affect the solution of a homogeneous differential equation?
4. Can the method of undetermined coefficients be used for non-linear differential equations?
5. How do you determine the form of the particular solution in the method of undetermined coefficients?
6. What happens if the non-homogeneous term is already part of the homogeneous solution?
7. What are some applications of second-order linear differential equations in physics?
8. How does the solution behavior change if the equation is \( y'' - 2y' + y = \sin(2x) \)?

**Tip:** When solving differential equations, always check if the form of your particular solution might overlap with the homogeneous solution; if it does, you’ll need to multiply by \( x \) to find a suitable particular solution.

Learn how to solve the differential equation y'' - 2y' + y = x + 1 + sin(2x) using the method of undetermined coefficients. This problem covers second-order linear non-homogeneous differential equations and includes step-by-step solutions and formula applications.

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