The given initial-value problem is:

$y'' + 16y = 0, \quad y(0) = 2, \quad y'(0) = -2$

Step 1: Solve the homogeneous differential equation

This is a second-order linear homogeneous differential equation with constant coefficients. We first find the characteristic equation associated with this differential equation:

$r^2 + 16 = 0$

Solving for $r$, we get:

$r^2 = -16 \quad \Rightarrow \quad r = \pm 4i$

Thus, the general solution to the homogeneous equation is:

$y(t) = C_1 \cos(4t) + C_2 \sin(4t)$

where $C_1$ and $C_2$ are constants to be determined by the initial conditions.

Step 2: Apply initial conditions

We are given the initial conditions:

1. $y(0) = 2$
2. $y'(0) = -2$

Apply $y(0) = 2$:

Substitute $t = 0$ into the general solution:

$y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1$

Thus, we have:

$C_1 = 2$

Apply $y'(0) = -2$:

To apply the second initial condition, we first find the derivative of $y(t)$:

$y'(t) = -4C_1 \sin(4t) + 4C_2 \cos(4t)$

Now substitute $t = 0$ into $y'(t)$:

$y'(0) = -4C_1 \sin(0) + 4C_2 \cos(0) = 4C_2$

We are given that $y'(0) = -2$, so:

$4C_2 = -2 \quad \Rightarrow \quad C_2 = -\frac{1}{2}$

Step 3: Write the particular solution

Substituting the values of $C_1$ and $C_2$ into the general solution:

$y(t) = 2 \cos(4t) - \frac{1}{2} \sin(4t)$

Thus, the solution to the initial-value problem is:

$y(t) = 2 \cos(4t) - \frac{1}{2} \sin(4t)$

Final Answer:

$y(t) = 2 \cos(4t) - \frac{1}{2} \sin(4t)$

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Would you like more details on any step or have any questions?

Here are some related questions you might find useful:

1. How would the solution change if the coefficient in front of $y$ was different from 16?
2. How do you solve second-order differential equations with non-homogeneous terms?
3. How can you determine the type of solution (oscillatory, exponential, etc.) from the characteristic equation?
4. What would happen if the initial conditions were modified to something like $y(0) = 0$ and $y'(0) = 1$?
5. Can we use the method of undetermined coefficients for this type of problem?

Tip: Always start by solving the characteristic equation for constant-coefficient linear differential equations, as it provides a direct route to the general solution!