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To solve the second-order nonhomogeneous differential equation:

\[
y'' - 2y' + 2y = x + 1
\]

with initial conditions \( y(0) = 3 \) and \( y'(0) = 0 \), we proceed in two main steps: 

### Step 1: Solve the Homogeneous Equation
The corresponding homogeneous equation is:
\[
y'' - 2y' + 2y = 0
\]

1. Write the characteristic equation:
   \[
   r^2 - 2r + 2 = 0
   \]
2. Solve this characteristic equation using the quadratic formula:
   \[
   r = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i
   \]

   The roots are complex: \( r = 1 \pm i \).

3. The general solution to the homogeneous equation is:
   \[
   y_h(x) = e^x (C_1 \cos x + C_2 \sin x)
   \]

### Step 2: Find a Particular Solution
For the nonhomogeneous term \( x + 1 \), we try a particular solution of the form:
\[
y_p(x) = Ax + B
\]

1. Substitute \( y_p(x) = Ax + B \) into the original equation \( y'' - 2y' + 2y = x + 1 \).

   - \( y_p' = A \)
   - \( y_p'' = 0 \)

   Substituting these into the equation gives:
   \[
   0 - 2A + 2(Ax + B) = x + 1
   \]
   Simplifying, we get:
   \[
   2Ax - 2A + 2B = x + 1
   \]

2. Equate coefficients for \( x \) and the constant term:
   - For \( x \): \( 2A = 1 \Rightarrow A = \frac{1}{2} \)
   - For the constant term: \( -2A + 2B = 1 \Rightarrow -1 + 2B = 1 \Rightarrow B = 1 \)

   Thus, the particular solution is:
   \[
   y_p(x) = \frac{1}{2}x + 1
   \]

### Step 3: General Solution
The general solution to the differential equation is:
\[
y(x) = y_h(x) + y_p(x) = e^x (C_1 \cos x + C_2 \sin x) + \frac{1}{2}x + 1
\]

### Step 4: Apply Initial Conditions
Use \( y(0) = 3 \) and \( y'(0) = 0 \) to find \( C_1 \) and \( C_2 \).

1. **At \( x = 0 \):**
   \[
   y(0) = e^0 (C_1 \cdot 1 + C_2 \cdot 0) + \frac{1}{2} \cdot 0 + 1 = C_1 + 1
   \]
   Since \( y(0) = 3 \), we get:
   \[
   C_1 + 1 = 3 \Rightarrow C_1 = 2
   \]

2. **For \( y'(0) = 0 \):**
   Differentiate \( y(x) \):
   \[
   y'(x) = e^x (C_1 \cos x + C_2 \sin x) + e^x (-C_1 \sin x + C_2 \cos x) + \frac{1}{2}
   \]
   At \( x = 0 \):
   \[
   y'(0) = e^0 (C_1 \cdot 0 + C_2 \cdot 1) + \frac{1}{2} = C_2 + \frac{1}{2}
   \]
   Since \( y'(0) = 0 \), we get:
   \[
   C_2 + \frac{1}{2} = 0 \Rightarrow C_2 = -\frac{1}{2}
   \]

### Final Solution
The solution to the differential equation is:
\[
y(x) = e^x \left( 2 \cos x - \frac{1}{2} \sin x \right) + \frac{1}{2}x + 1
\]

Would you like further details or have any questions?

Here are five related questions to explore:

1. What are the steps to solve a second-order nonhomogeneous differential equation in general?
2. How would the solution change if the initial conditions were different?
3. What is the significance of the characteristic equation in finding solutions to homogeneous differential equations?
4. How do complex roots in the characteristic equation affect the form of the solution?
5. Can we apply variation of parameters instead of the method of undetermined coefficients here?

**Tip:** For nonhomogeneous equations, always solve the homogeneous part first and then find the particular solution to simplify the process.

Find the solution of the differential equation y'' - 2y' + 2y = x + 1 with initial conditions y(0) = 3 and y'(0) = 0.

Learn how to solve the nonhomogeneous differential equation y'' - 2y' + 2y = x + 1 with initial conditions y(0) = 3 and y'(0) = 0. This solution uses the method of undetermined coefficients and covers finding the general solution with complex roots and applying initial conditions.

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