91Ó°ÊÓ

A non-homogeneous differential equation includes a term that is not solely dependent on the function and its derivatives. In this exercise, the equation is \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x = e^{-3 t} \). The term \( e^{-3 t} \) signifies its non-homogeneous nature because it stands independently from the variables and their derivatives.
In solving non-homogeneous equations, we seamlessly work with two parts:

• Homogeneous Equation: Here, we replace the independent term \( e^{-3 t} \) with \( 0 \) to get a homogeneous equation. Solving this helps us find the complementary solution.
• Particular Solution: We find this solution by determining an expression that satisfies the whole non-homogeneous equation. The particular solution addresses the independent term within the equation.

This layered approach—dealing with both homogenous and particular solutions—enables us to solve non-homogeneous differential equations efficiently.

To find a specific solution from the general solution of a differential equation, we utilize initial conditions. These conditions provide us crucial information about the function and its derivative at a specific point in time.
In our exercise, the initial conditions are given at \( t=0 \):

• \( x(0) = \frac{1}{2} \)
• \( \frac{\mathrm{d} x}{\mathrm{~d} t}(0) = -2 \)

These allow us to solve for the unknown constants present in the general solution. By substituting these conditions into the general solution, along with its derivative, we generate a system of equations. Solving these equations gives us the unique values of the constants which define the specific solution that meets the initial conditions.

This method is a common technique to find a particular solution to non-homogeneous linear differential equations. It involves making an educated guess about the form of the particular solution.
For example, in the equation given, we postulate a solution form like \( x_p(t) = Ae^{-3t} \). Why go for this form? It's because the non-homogeneous term \( e^{-3t} \) suggests that the particular solution should mimic this appearance.
Once you have a guess, substitute it back into the differential equation. This allows you to determine the unknown coefficient (here \( A \)). After substitution and simplification, we found \( A = -\frac{1}{6} \). This substitution process ensures the particular solution addresses the non-homogeneous aspect of the differential equation.

The concept of a characteristic equation is instrumental in solving homogeneous linear differential equations. It derives from assuming a solution of exponential form for the homogeneous part of a differential equation.
In our case, we have \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} + 4 \frac{\mathrm{d} x}{\mathrm{~d} t} + 3x = 0 \). We assume a solution of the form \( x = e^{rt} \), substituting this leads to the characteristic equation \( r^2 + 4r + 3 = 0 \).
Solving this quadratic equation by factoring gives us roots \( r = -1 \) and \( r = -3 \). These roots define the general solution to the homogeneous equation:

• \( x_h(t) = C_1 e^{-t} + C_2 e^{-3t} \)

Using the characteristic equation ensures the solutions perfectly fit the structure of the original homogeneous differential equation, providing its general solution form.