91Ó°ÊÓ

The homogeneous solution of a differential equation provides insight into the system's natural response when there's no external input. For the equation given in the exercise, we begin by solving the homogeneous equation \(y'' + y = 0\). The method involves assuming a solution of the form \(y_h = e^{rt}\), where \(r\) is a constant to be determined. By substituting \(y_h\) into the differential equation, we arrive at the characteristic equation, which is crucial for finding the roots \(r\). We get:

• \(r^2 + 1 = 0\)

Solving this characteristic equation, we find the roots to be \(r = \pm i\). These roots suggest oscillatory solutions because they are complex. The homogeneous solution is then:

• \(y_h(t) = C_1\cos(t) + C_2\sin(t)\)

This solution represents the natural behavior of the system without external forces.

The particular solution accounts for the non-homogeneous part of the differential equation, which stems from external inputs or forces. Given the equation \(y'' + y = \tan t + e^{3t} - 1\), our task is to find a function \(y_p\) that satisfies this equation. We make an educated guess, often referred to as the method of undetermined coefficients. We assume a particular solution of the form:

• \(y_p(t) = A \tan(t) + B e^{3t} + C\)

Substituting \(y_p(t)\) and its derivatives into the non-homogeneous differential equation allows us to determine the constants \(A\), \(B\), and \(C\) by equating coefficients of similar terms. This process can be intricate due to the non-standard functions involved, but it is essential for capturing the influence of the non-homogeneous terms.

The general solution of a non-homogeneous differential equation combines both the homogeneous and particular solutions. It encompasses the entire behavior of the system, influenced by both its natural dynamics and external factors. For our problem, the general solution \(y(t)\) is expressed as:

• \(y(t) = y_h(t) + y_p(t)\)
• \(y(t) = C_1\cos(t) + C_2\sin(t) + A\tan(t) + Be^{3t} + C\)

This expression satisfies both the homogeneous part of the equation and the non-homogeneous influences \(\tan t + e^{3t} - 1\). By tweaking the constants \(C_1\), \(C_2\), \(A\), \(B\), and \(C\), one can adapt the solution to various initial conditions, giving it versatility across different contexts.

The characteristic equation arises from the assumption of exponential solutions in linear differential equations. It is derived by inserting our assumed solution \(y_h = e^{rt}\) into the homogeneous version of the equation, leading to a polynomial that reflects the dynamic nature of the system. For the equation in the exercise, the characteristic equation is:

• \(r^2 + 1 = 0\)

This result tells us about the oscillatory nature of the system, as the roots \(r = \pm i\) indicate sinusoidal solutions (sines and cosines). The characteristic equation is a powerful tool for understanding the intrinsic responses of the system, guiding the formulation of the homogeneous solution.