91Ó°ÊÓ

Verification of a solution in the context of differential equations is a crucial step. It confirms that a given function satisfies the equation. Here, we are verifying if the function \( y(t) = C_{1} \sin{4t} + C_{2} \cos{4t} \) solves the differential equation \( y''(t) + 16y(t) = 0 \).
Verification involves a few steps. First, compute the derivatives of the given function. Then, substitute these derivatives into the original differential equation. Finally, simplify the expression. If the left-hand side equals the right-hand side, the function is a solution.
This process ensures that the function holds true for all values of \( t \) and arbitrary constants \( C_{1} \) and \( C_{2} \), confirming the function's validity as a general solution.

To solve differential equations, understanding derivatives is key, especially higher order ones. Here, calculating the second derivative is necessary as it features in the given differential equation.
The function provided is \( y(t) = C_{1} \sin{4t} + C_{2} \cos{4t} \). The process involves finding the first derivative \( y'(t) \) first:

• The derivative of \( C_{1} \sin{4t} \) is \( 4C_{1} \cos{4t} \).
• The derivative of \( C_{2} \cos{4t} \) is \(-4C_{2} \sin{4t} \).

So, \( y'(t) = 4C_{1} \cos{4t} - 4C_{2} \sin{4t} \).
Next, derive \( y'(t) \) to find \( y''(t) \):

• The derivative of \( 4C_{1} \cos{4t} \) is \(-16C_{1} \sin{4t} \).
• The derivative of \(-4C_{2} \sin{4t} \) is \(-16C_{2} \cos{4t} \).

Thus, \( y''(t) = -16C_{1}\sin{4t} - 16C_{2} \cos{4t} \).
This step-by-step approach is essential when working with differential equations.

In this problem, trigonometric functions \( \sin \) and \( \cos \) are used within \( y(t) \). These functions are fundamental in mathematics and appear often in differential equations.
Trigonometric functions have specific derivatives that we used here:

• The derivative of \( \sin{x} \) is \( \cos{x} \).
• The derivative of \( \cos{x} \) is \( -\sin{x} \).

Moreover, these functions feature periodic properties, making them invaluable within physics and engineering contexts.
Understanding their behavior and derivations not only aids in solving equations like the one in this problem but also paves the way for tackling more complex scenarios in mathematical modeling.