The given differential equation is:

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

Here, \(D\) represents the differentiation operator (i.e., \(D = \frac{d}{dx}\)). So, \((D-1)^2 y\) can be expanded and understood as applying the operator \((D-1)\) twice to \(y\).

Let's proceed step by step:

### 1. Expanding the operator \((D-1)^2 y\):
The operator \((D - 1)\) means \(D - 1 = \frac{d}{dx} - 1\). Now, applying this operator twice:

\[
(D-1)^2 y = \left(\frac{d}{dx} - 1\right)\left(\frac{d}{dx} - 1\right) y
\]

Expanding the product:

\[
= \left(\frac{d}{dx} - 1\right)\left(\frac{dy}{dx} - y\right)
\]
\[
= \frac{d^2y}{dx^2} - \frac{dy}{dx} - \frac{dy}{dx} + y
\]
\[
= \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y
\]

Thus, the operator \((D-1)^2 y\) becomes:

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

### 2. Given equation:
The equation now becomes:

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

### 3. Solving the homogeneous part:
First, solve the homogeneous equation:

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

The characteristic equation is:

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

This simplifies to:

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

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

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

### 4. Solving the non-homogeneous part:
The non-homogeneous equation is:

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

We use the method of undetermined coefficients to solve this. The right-hand side contains \(e^x\) and a polynomial \(x\), so we propose a particular solution of the form:

\[
y_p = A x e^x + B x + C
\]

Now, differentiate \(y_p\):

\[
y_p' = A(e^x + x e^x) + B
\]
\[
y_p'' = A(2e^x + x e^x)
\]

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original equation \(y'' - 2y' + y = e^x + x\) and solve for \(A\), \(B\), and \(C\).

### 5. Final solution:
The general solution will be the sum of the homogeneous solution \(y_h\) and the particular solution \(y_p\):

\[
y = y_h + y_p
\]

This gives:

\[
y = (C_1 + C_2 x) e^x + (A x e^x + B x + C)
\]

Would you like to proceed with finding the values of \(A\), \(B\), and \(C\), or need further clarification?

### Related Questions:
1. What are the initial conditions if they were provided?
2. Can we use variation of parameters to solve this instead of undetermined coefficients?
3. How does the solution behave as \(x \to \infty\)?
4. What happens if the right-hand side was a different function, such as \(e^{2x}\)?
5. Can this method apply to higher-order differential equations?

### Tip:
Always verify if the particular solution you assume satisfies the original differential equation to ensure accuracy.

Explore the solution to the differential equation y'' - 2y' + y = e^x + x using the method of undetermined coefficients. Learn step-by-step how to solve both the homogeneous and non-homogeneous parts of the equation, including the characteristic equation and general solution. Suitable for undergraduate students studying differential equations.

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