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The problem asks to solve the second-order differential equation: 

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

This is a homogeneous linear differential equation with constant coefficients.

### Step 1: Write the characteristic equation.
For a second-order differential equation of the form:
\[
y'' + by' + cy = 0
\]
The characteristic equation is:
\[
r^2 + br + c = 0
\]

For our equation \( y'' - 4y' + 4y = 0 \), the characteristic equation becomes:
\[
r^2 - 4r + 4 = 0
\]

### Step 2: Solve the characteristic equation.
\[
r^2 - 4r + 4 = (r - 2)^2 = 0
\]
So, the characteristic root is \( r = 2 \), with multiplicity 2.

### Step 3: Write the general solution.
When you have a repeated root \( r = 2 \), the general solution is:
\[
y(t) = (C_1 + C_2t)e^{2t}
\]
where \( C_1 \) and \( C_2 \) are constants.

Thus, the solution is:
\[
y(t) = (C_1 + C_2t)e^{2t}
\]

### Answer
The correct answer is:

**Option 4: \( y = (C_1 + C_2t)e^{2t} \)**

---

Would you like further explanation on any of the steps?

Here are 5 related questions you might find useful:
1. How do you solve non-homogeneous second-order differential equations?
2. What is the solution when the characteristic equation has complex roots?
3. How do boundary conditions affect the solution of a differential equation?
4. Can you explain the method of undetermined coefficients?
5. What is the difference between particular and general solutions in differential equations?

**Tip:** When solving higher-order differential equations, always first solve the characteristic equation to identify the nature of the solution!

Solve the second-order differential equation: y'' - 4y' + 4y = 0

Learn how to solve the second-order differential equation y'' - 4y' + 4y = 0 using characteristic equations. This problem covers differential equations concepts and provides a detailed step-by-step solution for the homogeneous linear case.

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