We are given the differential equation:

\[
y'' - 2y' + y = (D - 1)^2 y = e^x + x
\]

Where \( D \) is the differentiation operator, meaning \( D y = y' \) and \( D^2 y = y'' \). We can interpret \( (D - 1)^2 \) as the differential operator applied twice, and rewrite the equation to solve it step by step.

### Step 1: Expand the operator
We first expand \( (D - 1)^2 \):

\[
(D - 1)^2 = (D - 1)(D - 1) = D^2 - 2D + 1
\]

So, we have:

\[
(D^2 - 2D + 1)y = e^x + x
\]

This is the same as:

\[
y'' - 2y' + y = e^x + x
\]

### Step 2: Solve the homogeneous equation
The homogeneous part of the equation is:

\[
y'' - 2y' + y = 0
\]

The characteristic equation for this is:

\[
r^2 - 2r + 1 = 0
\]

Factoring gives:

\[
(r - 1)^2 = 0
\]

So, \( r = 1 \) is a repeated root. Therefore, the general solution to the homogeneous equation is:

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

### Step 3: Solve the particular equation
We now solve the non-homogeneous equation:

\[
y'' - 2y' + y = e^x + x
\]

For the right-hand side \( e^x + x \), we try a particular solution of the form:

\[
y_p(x) = A x e^x + Bx + C
\]

We will find \( A \), \( B \), and \( C \) by substituting this guess into the differential equation.

### Step 4: Find the particular solution
1. Compute \( y_p' \) and \( y_p'' \):
   - \( y_p'(x) = A e^x + A x e^x + B \)
   - \( y_p''(x) = 2A e^x + A x e^x \)

2. Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the differential equation:

\[
(2A e^x + A x e^x) - 2(A e^x + A x e^x + B) + (A x e^x + Bx + C) = e^x + x
\]

Collect like terms:

\[
(A x e^x - 2A x e^x + A x e^x) + (2A e^x - 2A e^x) + (-2B + Bx + C) = e^x + x
\]

Simplify:

\[
0 + (-2B + Bx + C) = e^x + x
\]

This gives the following system of equations:

\[
B = 1, \quad -2B + C = 0
\]

Substituting \( B = 1 \) into the second equation:

\[
-2(1) + C = 0 \quad \Rightarrow \quad C = 2
\]

### Step 5: General solution
The particular solution is:

\[
y_p(x) = x + 2
\]

Thus, the general solution to the differential equation is:

\[
y(x) = (C_1 + C_2 x) e^x + x + 2
\]

### Do you have any questions, or would you like more details?

### Related questions:
1. How do we find the particular solution for more complicated non-homogeneous terms?
2. What does it mean to have a repeated root in the characteristic equation?
3. How do we apply the method of undetermined coefficients for different types of non-homogeneous terms?
4. How would the solution change if the right-hand side involved trigonometric functions instead of \( e^x + x \)?
5. What are the steps to solve a second-order differential equation with variable coefficients?

### Tip:
When solving non-homogeneous differential equations, choosing an appropriate form for the particular solution is key.

Explore the step-by-step solution of the second-order differential equation y'' - 2y' + y = e^x + x. This problem involves expanding operators, solving the characteristic equation, and finding both homogeneous and particular solutions. Ideal for Grades 11-12 and early university students.

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