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To find the solution of the given differential equation with initial conditions, let's solve it step by step.

The differential equation is:

\[ y'' + 8y' = 512 \sin(8t) + 1152 \cos(8t) \]

with initial conditions:

\[ y(0) = 4 \]
\[ y'(0) = 5 \]

### Step 1: Solve the homogeneous equation
The homogeneous part of the equation is:

\[ y'' + 8y' = 0 \]

The characteristic equation for this differential equation is:

\[ r^2 + 8r = 0 \]
\[ r(r + 8) = 0 \]

This gives us the roots:

\[ r = 0 \]
\[ r = -8 \]

So, the general solution of the homogeneous equation is:

\[ y_h(t) = C_1 + C_2 e^{-8t} \]

### Step 2: Find a particular solution
For the non-homogeneous equation:

\[ y'' + 8y' = 512 \sin(8t) + 1152 \cos(8t) \]

We use the method of undetermined coefficients. Assume a particular solution of the form:

\[ y_p(t) = A \sin(8t) + B \cos(8t) \]

Compute the first and second derivatives of \( y_p(t) \):

\[ y_p'(t) = 8A \cos(8t) - 8B \sin(8t) \]
\[ y_p''(t) = -64A \sin(8t) - 64B \cos(8t) \]

Substitute \( y_p(t) \), \( y_p'(t) \), and \( y_p''(t) \) into the non-homogeneous differential equation:

\[ (-64A \sin(8t) - 64B \cos(8t)) + 8(8A \cos(8t) - 8B \sin(8t)) = 512 \sin(8t) + 1152 \cos(8t) \]
\[ -64A \sin(8t) - 64B \cos(8t) + 64A \cos(8t) - 64B \sin(8t) = 512 \sin(8t) + 1152 \cos(8t) \]
\[ (-64B - 64B) \sin(8t) + (64A - 64A) \cos(8t) = 512 \sin(8t) + 1152 \cos(8t) \]
\[ -128B \sin(8t) + 0 \cos(8t) = 512 \sin(8t) + 1152 \cos(8t) \]

This simplifies to:

\[ -128B \sin(8t) = 512 \sin(8t) \]
\[ 0 = 1152 \cos(8t) \]

From this, we find:

\[ -128B = 512 \]
\[ B = -4 \]

\[ 64A = 1152 \]
\[ A = 18 \]

Thus, the particular solution is:

\[ y_p(t) = 18 \sin(8t) - 4 \cos(8t) \]

### Step 3: Combine solutions and apply initial conditions
The general solution of the differential equation is:

\[ y(t) = y_h(t) + y_p(t) \]
\[ y(t) = C_1 + C_2 e^{-8t} + 18 \sin(8t) - 4 \cos(8t) \]

Now apply the initial conditions to determine \( C_1 \) and \( C_2 \):

1. When \( t = 0 \):

\[ y(0) = C_1 + C_2 e^0 + 18 \sin(0) - 4 \cos(0) = 4 \]
\[ C_1 + C_2 - 4 = 4 \]
\[ C_1 + C_2 = 8 \]

2. When \( t = 0 \):

\[ y'(t) = -8C_2 e^{-8t} + 144 \cos(8t) + 32 \sin(8t) \]
\[ y'(0) = -8C_2 + 144 \cos(0) + 32 \sin(0) = 5 \]
\[ -8C_2 + 144 = 5 \]
\[ -8C_2 = -139 \]
\[ C_2 = \frac{139}{8} = 17.375 \]

Substitute \( C_2 \) back into the first initial condition equation:

\[ C_1 + 17.375 = 8 \]
\[ C_1 = 8 - 17.375 \]
\[ C_1 = -9.375 \]

Thus, the final solution is:

\[ y(t) = -9.375 + 17.375 e^{-8t} + 18 \sin(8t) - 4 \cos(8t) \]

Do you want details or have any questions about this solution?

Here are 5 related questions:
1. How do you find the particular solution for non-homogeneous differential equations using the method of undetermined coefficients?
2. What is the significance of the characteristic equation in solving homogeneous differential equations?
3. How can initial conditions be used to find specific values for constants in the general solution?
4. What are the differences between the complementary function and particular solution in differential equations?
5. How does the presence of trigonometric functions in the non-homogeneous term affect the form of the particular solution?

**Tip:** Always verify your solution by substituting it back into the original differential equation to ensure it satisfies both the equation and the initial conditions.

Find the solution of y'' + 8y' = 512 sin(8t) + 1152 cos(8t) with initial conditions y(0) = 4 and y'(0) = 5.

Learn how to solve the second-order differential equation y'' + 8y' = 512 sin(8t) + 1152 cos(8t) with initial conditions y(0) = 4 and y'(0) = 5. This problem explores key concepts in differential equations, including homogeneous solutions and the method of undetermined coefficients. Suitable for advanced high school calculus or undergraduate mathematics students.

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