To solve the differential equation

\[
y'' + 16y = 0,
\]

with the initial conditions \( y(0) = 2 \) and \( y'(0) = -2 \), we proceed as follows:

### Step 1: Solve the Characteristic Equation

The given equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is obtained by assuming a solution of the form \( y = e^{rt} \), leading to:

\[
r^2 + 16 = 0.
\]

Solving this, we get:

\[
r^2 = -16 \implies r = \pm 4i.
\]

So, the general solution to the differential equation is:

\[
y(t) = C_1 \cos(4t) + C_2 \sin(4t).
\]

### Step 2: Apply Initial Conditions

1. **Using \( y(0) = 2 \):**

   Substitute \( t = 0 \):

   \[
   y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 2.
   \]

   So, \( C_1 = 2 \).

2. **Using \( y'(0) = -2 \):**

   First, compute the derivative of \( y(t) \):

   \[
   y'(t) = -4C_1 \sin(4t) + 4C_2 \cos(4t).
   \]

   Substitute \( t = 0 \):

   \[
   y'(0) = -4C_1 \sin(0) + 4C_2 \cos(0) = 4C_2 = -2.
   \]

   So, \( C_2 = -\frac{1}{2} \).

### Step 3: Write the Particular Solution

The particular solution that satisfies the given initial conditions is:

\[
y(t) = 2 \cos(4t) - \frac{1}{2} \sin(4t).
\]

### Final Answer:

\[
\boxed{y(t) = 2 \cos(4t) - \frac{1}{2} \sin(4t).}
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you solve second-order linear homogeneous differential equations with different characteristic roots?
2. What is the role of initial conditions in determining a unique solution for a differential equation?
3. How would the solution change if the initial conditions were different?
4. What is the significance of complex roots in the characteristic equation?
5. Can you explain the physical meaning of cosine and sine terms in solutions of differential equations?

**Tip:** When dealing with trigonometric solutions, use initial conditions to determine coefficients accurately, ensuring the solution fits the specific problem context.

The query submitted by the user: y''+16y=0, y(0)=2, y'(0)=-2

Learn how to solve the second-order linear homogeneous differential equation y'' + 16y = 0 with initial conditions y(0) = 2 and y'(0) = -2. This problem involves solving the characteristic equation and applying initial values to find a particular solution. Suitable for advanced undergraduate mathematics students.

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