Solve the following differential equations. \(\left(D^{2}+16\right) y=0\)

A linear homogeneous differential equation is a type of differential equation where the function and its derivatives appear linearly, and all terms involve the dependent variable or its derivatives. In simple terms, it doesn’t have any isolated constants or functions of the independent variable only.

For example, consider the equation \((D^{2}+16) y = 0\). This is a second-order linear homogeneous differential equation. In this case, everything apart from the dependent variable y depends on D (differentiation operator), and no standalone terms are present.

The term 'homogeneous' means that all terms involve y or its derivatives, without any constant or external forcing function. If you see an equation like this and there are no extra terms, it’s homogeneous.

To solve a second-order linear homogeneous differential equation, we start by forming the characteristic equation. This is a crucial step that simplifies the process.

A characteristic equation is obtained by replacing the differentiation operator D with a variable, usually denoted by r. So for the given differential equation \((D^{2}+16) y=0\), the characteristic equation is formed as follows:

\[D^2 + 16 = 0 \]

This now translates to:

\[r^2 + 16 = 0\]

This quadratic equation replaces the original differential equation, making it easier to solve.

Often, the characteristic equation can yield complex roots, which are vital for solving differential equations.

If we solve the characteristic equation \({r^2 + 16 = 0})\), we get:

\[ r^2 = -16 \]

Taking the square root of both sides, we find:

\[ r = \pm 4i \]

These complex roots emerge when the characteristic equation has no real solutions. In such cases, the imaginary unit i (where i is defined as the square root of -1) helps in representing these complex roots.
An example of the roots in this scenario are \(\pm 4i\), which denote both positive and negative imaginary numbers.

With complex roots, the general solution of the differential equation combines trigonometric functions.

For the given roots \(\pm 4i\), the solution includes cosine and sine components. These are tied together with arbitrary constants to complete the solution.

The general solution is given by:

\[ y(t) = C_{1} cos(4t) + C_{2} sin(4t) \]

Here, \(C_{1}\) and \(C_{2}\) are arbitrary constants that are determined by initial conditions or boundary conditions, which can be specific values or constraints provided in the problem.