91Ó°ÊÓ

In the context of differential equations, the homogeneous solution is a key component when dealing with non-homogeneous linear differential equations. For our given problem, the homogeneous differential equation is derived by setting the right-hand side to zero: \( x''(t) + 4x(t) = 0 \).To find the homogeneous solution, we assume a solution of the form \( x_h(t) = e^{rt} \), where \( r \) is a constant. This assumption leads us to the concept of the characteristic equation: \( r^2 + 4 = 0 \).By solving the characteristic equation, we find that \( r = \pm 2i \). This indicates a solution involving sinusoidal functions, specifically given by the formula:

• \( x_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \)

These coefficients, \( C_1 \) and \( C_2 \), are constants determined by initial or boundary conditions. In this case, they are placeholders that will be used to satisfy additional criteria given by initial conditions later in the problem. Consequently, the homogeneous solution is foundational, as it provides the basis onto which a particular solution is added to address the non-homogeneity of the differential equation.

Finding a particular solution to a non-homogeneous differential equation involves crafting a specific function that satisfies the entire equation, including its non-homogeneous part. In the problem given, the differential equation is \( x''(t) + 4x(t) = -8t^2 \).The term \(-8t^2\) suggests that a polynomial solution might be suitable. Accordingly, we assume a particular solution of the form \( x_p(t) = At^2 + Bt + C \), which allows us to include terms up to the second degree, matching the degree of \(-8t^2\).By substituting \( x_p(t) \) into the differential equation, and comparing coefficients for powers of \( t \), we solve for \( A, B, \text{and } C \) to find that \( A = 2, B = 0, \text{and } C = 0 \). Thus, the particular solution becomes:

• \( x_p(t) = 2t^2 \)

This particular solution is specific to addressing the non-homogeneous component \(-8t^2\), satisfying the overall differential equation when added to the homogeneous solution.

The concept of a characteristic equation is central when solving linear homogeneous differential equations, especially those with constant coefficients. For our differential equation \( x''(t) + 4x(t) = 0 \), the characteristic equation is derived simply by assuming the solution takes on an exponential form \( x(t) = e^{rt} \), thus transforming the differential equation into an algebraic one.This translates to:

• \( r^2 + 4 = 0 \)

The characteristic equation \( r^2 + 4 = 0 \) is solved for \( r \), revealing complex roots \( r = \pm 2i \). These complex roots lead to solutions involving sine and cosine functions due to Euler's formula, which states \( e^{ix} = \cos(x) + i\sin(x) \).Thus, the resultant homogeneous solution, constructed using these roots, is \( x_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \). The characteristic equation effectively connects the differential equation to the form of its general solution, offering insights into the behavior of its solutions whether oscillatory, exponential, or a combination.

Initial conditions provide additional information to uniquely determine the constants in the general solution of a differential equation. In this problem, they are essential to finding the specific solution to: \( x(t) = C_1 \cos(2t) + C_2 \sin(2t) + 2t^2 \).We are given:

• \( x(0) = 3 \)
• \( x\left(\frac{1}{4}\pi\right) = 0 \)

For \( x(0) = 3 \), substituting \( t = 0 \) results in:\( 3 \cos(0) + C_2 \sin(0) + 2(0)^2 = 3 \). This equation simplifies to \( C_1 = 3 \), as sine of zero is zero and cosine of zero is one.For \( x\left(\frac{1}{4}\pi\right) = 0 \), substituting \( t = \frac{1}{4}\pi \) yields:\( 3 \cos\left(\frac{1}{2}\pi\right) - 1.5\pi \sin\left(\frac{1}{2}\pi\right) + 2\left(\frac{1}{4}\pi\right)^2 = 0 \). Solving, we find that \( C_2 = -1.5\pi \).These steps allow us to determine the constants and thereby finalize the specific solution, which completely satisfies both the differential equation and the initial conditions. Initial conditions are crucial in ensuring that the solution is not just any solution, but the exact one necessary for the scenario described.