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Understanding the characteristic equation is vital when dealing with second-order linear homogeneous differential equations. It allows us to find the solution for the function in the differential equation. To derive the characteristic equation from our differential equation, \(f''(t) + f'(t) - 12f(t) = 0\), we assume a solution of the form \(e^{rt}\), where \(r\) is a constant. This assumption standardizes the terms of the differential equation, leading to a quadratic equation in terms of \(r\). Here, the characteristic equation is \(r^2 + r - 12 = 0\). This approach turns the complex problem of solving the differential equation into a more familiar task—solving a quadratic equation.

The second-order linear homogeneous differential equations have a standard form consisting of the second derivative of a function plus a first derivative and the function itself, all set to zero. In our example, \(f''(t) + f'(t) - 12f(t) = 0\), demonstrates this form perfectly. The linearity implies that the equation's solutions can be superimposed to form new solutions, while homogeneity means that if \(f(t)\) is a solution, then so is any constant multiple of \(f(t)\). A key property of such equations is that their general solution can often be expressed as a combination of exponential functions.

A general solution to a differential equation encompasses all possible solutions. It includes arbitrary constants because it represents a family of solutions rather than a single specific solution. For our exercise, once the characteristic equation \(r^2 + r - 12 = 0\) has been solved and the roots \(r_1\) and \(r_2\) obtained as 3 and -4 respectively, we form the general solution by combining the exponential solutions associated with these roots: \(f(t) = C_1e^{3t} + C_2e^{-4t}\). The next crucial step involves determining the arbitrary constants \(C_1\) and \(C_2\), which is done by applying initial conditions.

Quadratic equations, such as those forming the characteristic equation of a differential problem, can be solved using various methods, including factoring, completing the square, or using the quadratic formula. Solving these equations provides the roots that are essential to formulating the general solution of the differential equation. In the given problem, factoring the characteristic equation gives us two roots, \(r_1 = 3\) and \(r_2 = -4\). This is an important step that bridges the original differential equation to its general solution.

Initial conditions are used to find the specific solution to a differential equation that not only satisfies the equation but also conforms to certain criteria at the outset. They are the key to determining the values of the arbitrary constants in the general solution. Applying the initial conditions \(f(0)=0\) and \(f'(0)=0\) to our general solution, we set up a system of equations to solve for \(C_1\) and \(C_2\). However, in our exercise, the unique solution to the system of equations yields \(C_1 = 0\) and \(C_2 = 0\), resulting in the trivial solution \(f(t) = 0\), which indicates that the function remains at zero for all \(t\). This outcome is consistent with the scenario where the initial state and rate of change are both zero.