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A second-order differential equation involves a second derivative in its formulation. These equations often appear in physical applications, such as systems involving motion. The general form of a second-order linear differential equation is \(a y'' + b y' + c y = f(t)\), where the highest derivative is of the second order. For the given problem, \(y'' - 16y = 16 te^{-4t}\), the equation involves the second derivative \(y''\) and no first derivative \(y'\). The term \(16 te^{-4t}\) makes the equation non-homogeneous.
To solve such equations, we split the process into two parts: solving the associated homogeneous equation, where \(f(t) = 0\), and finding a particular solution to the non-homogeneous equation.

The method of undetermined coefficients is a technique for finding the particular solution of a non-homogeneous differential equation. This method works well when the non-homogeneous term is of a specific form, such as exponentials, sines, cosines, or polynomials. For our problem, the non-homogeneous term is \(16t e^{-4t}\). We assume a form for \(y_p(t)\), the particular solution, based on this term.
Here are the steps:

• Identify the form of the non-homogeneous term
• Guess a particular solution with undetermined coefficients
• Determine the first and second derivatives of the guess
• Substitute these into the original differential equation
• Solve for the coefficients by comparing terms

In our case, we chose \(y_p(t) = A t e^{-4t} + B t^2 e^{-4t}\). After substituting into the equation and simplifying, we solved for \(A\) and \(B\). The solution was \(y_p(t) = -t e^{-4t}\).

Initial conditions provide specific values for the function and its derivatives at a certain point. These conditions are crucial for finding the particular solution that fits the given scenario. For our differential equation \(y'' - 16y = 16 te^{-4t}\), the initial conditions given are \(y(0) = 0\) and \(y'(0) = 0\).
Using these initial conditions in the general solution, \(y(t) = C_1 e^{4t} + C_2 e^{-4t} - t e^{-4t}\), we determine the specific values for the constants \(C_1\) and \(C_2\). Plugging \(t = 0\) into \(y(t)\) gives \(C_1 + C_2 = 0\). Taking the derivative \(y'(t) = 4C_1 e^{4t} - 4C_2 e^{-4t} - e^{-4t}(1 - 4t)\) and substituting \(t = 0\) gives \(4C_1 - 4C_2 - 1 = 0\). Solving these simultaneous equations, we find \(C_1 = \frac{1}{8}\) and \(C_2 = -\frac{1}{8}\). Thus, the final particular solution is \[y(t) = \frac{1}{8} e^{4t} - \frac{1}{8} e^{-4t} - t e^{-4t}\].