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To solve the differential equation

\[
y'' - 2y' - 8y = -16t + 16t^2,
\]

we'll approach it by finding the general solution, which consists of two parts: the complementary solution \( y_c \) and the particular solution \( y_p \).

### Step 1: Find the Complementary Solution \( y_c \)
The complementary solution solves the homogeneous equation:
\[
y'' - 2y' - 8y = 0.
\]

1. Start by finding the characteristic equation:
   \[
   r^2 - 2r - 8 = 0.
   \]
2. Solve this quadratic equation using the quadratic formula:
   \[
   r = \frac{2 \pm \sqrt{(-2)^2 + 4 \cdot 8}}{2} = \frac{2 \pm 6}{2}.
   \]
   This gives roots \( r = 4 \) and \( r = -2 \).

Since both roots are real and distinct, the complementary solution is:
\[
y_c = c_1 e^{4t} + c_2 e^{-2t}.
\]

### Step 2: Find the Particular Solution \( y_p \)
Since the non-homogeneous part is \( -16t + 16t^2 \), a polynomial of degree 2, we assume a particular solution of the form:
\[
y_p = At^2 + Bt + C.
\]

1. Find \( y_p' \) and \( y_p'' \):
   - \( y_p' = 2At + B \),
   - \( y_p'' = 2A \).

2. Substitute \( y_p \), \( y_p' \), and \( y_p'' \) into the differential equation:
   \[
   2A - 2(2At + B) - 8(At^2 + Bt + C) = -16t + 16t^2.
   \]

3. Collect terms and equate coefficients to solve for \( A \), \( B \), and \( C \).

### Step

Find the general solution of the differential equation y'' - 2y' - 8y = -16t + 16t^2.

This guide provides a step-by-step solution to the differential equation y'' - 2y' - 8y = -16t + 16t^2. It covers finding the complementary solution using the characteristic equation and the particular solution with polynomial guessing. Suitable for advanced high school or college-level students.

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