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When solving differential equations, the initial conditions play a crucial role in determining the specific solution from the general family of solutions. They provide information about the value of the function and possibly its derivatives at a particular point in time or space. In this context, we have two initial conditions given:

• \(f(0) = 0\) means when \(t = 0\), the displacement \(y\) is zero. This typically reflects the state of the object at the start of the observation or experiment.
• \(f(1) = 0.50 \text{ cm}\) indicates that when \(t = 1\), the displacement \(y\) is 0.5 cm. This gives us a specific future behavior at a certain time point.

These conditions help us find specific values for the constants \(C_1\) and \(C_2\) in the general solution. Without initial conditions, we would have only a general solution that involves arbitrary constants. By applying these conditions, we transition from a general solution to a particular solution that fits the described physical situation.

The characteristic equation is a vital part of solving second-order linear homogeneous differential equations. We start by assuming a trial solution of the form \(y = e^{rt}\), where \(r\) is a constant we need to find. This assumption transforms the differential equation into an algebraic equation where derivatives of \(e^{rt}\) are involved.To get the characteristic equation from our differential equation \(\frac{d^2 y}{dt^2} + 4 \frac{dy}{dt} + 4y = 0\), substitute the trial solution into the equation and simplify. This yields the characteristic equation:\[ r^2 + 4r + 4 = 0 \]This is a quadratic equation, and solving it provides the roots necessary to formulate the general solution. Here, the roots are repeated with \(r = -2\). This indicates a particular form of the general solution associated with repeated roots, which will be discussed in the next section.

The general solution of a second-order linear homogeneous differential equation depends on the roots of the characteristic equation. For the equation \(\frac{d^2 y}{dt^2} + 4 \frac{dy}{dt} + 4y = 0\), the characteristic equation gives a repeated root \(r = -2\). When roots are repeated, as with our root, the general solution is expressed as:\[ y(t) = (C_1 + C_2 t) e^{-2t} \]where \(C_1\) and \(C_2\) are constants determined by the initial conditions. If the roots had been distinct, the solution would involve terms of the form \(C_1 e^{r_1 t} + C_2 e^{r_2 t}\), but repeated roots lead to this slightly more complex form involving \(t\) in one of the terms. The initial conditions \(f(0) = 0\) and \(f(1) = 0.50\) are then used to solve for \(C_1\) and \(C_2\), resulting in a particular solution that reflects the physical properties and conditions of the problem described.