91Ó°ÊÓ

The characteristic equation is a crucial concept in solving ordinary differential equations (ODEs). It provides the key to finding a general solution. For a linear differential equation with constant coefficients, we can assume a solution of the form \( y = e^{rt} \), which helps reduce the differential equation to an algebraic form.
This algebraic form is known as the characteristic equation. In our example from the exercise, we start with the third-order differential equation \( y''' - 4y' = 0 \). By substituting \( y = e^{rt} \), we derive the characteristic equation \( r^3 - 4r = 0 \). This is done by calculating derivatives \( y', y'', y''' \) and then replacing them in the ODE.
After performing algebraic manipulations, such as factoring, we find the roots: \( r = 0 \), \( r = 2 \), and \( r = -2 \). These roots are vital as they determine the form of the linearly independent solutions.

Once the roots of the characteristic equation are known, we can write the general solution for the differential equation. The roots provide us with a set of exponential functions, or perhaps polynomial functions, when the roots are repeated, that serve as the building blocks of the solution.
In our exercise, the distinct roots \( r = 0, 2, -2 \) lead to the solutions: \( y_1 = 1 \), \( y_2 = e^{2t} \), and \( y_3 = e^{-2t} \). These solutions are linearly independent because none of them can be expressed as a linear combination of the others.
To check linear independence, construct a set of solutions from the characteristic roots. If the Wronskian, calculated from these solutions, does not vanish, the solutions are linearly independent. This means that any general solution of the ODE can be represented as a combination of these linearly independent functions.

The Wronskian is a determinant used to determine the linear independence of a set of functions. Calculating the Wronskian involves creating a matrix using the functions and their derivatives. For our three solutions \( y_1 = 1 \), \( y_2 = e^{2t} \), and \( y_3 = e^{-2t} \), the Wronskian matrix is constructed by arranging the functions in the first row, their first derivatives in the second, and their second derivatives in the third.
The matrix looks like this:
\[ W(y_1, y_2, y_3) = \begin{vmatrix} 1 & e^{2t} & e^{-2t} \ 0 & 2e^{2t} & -2e^{-2t} \ 0 & 4e^{2t} & 4e^{-2t} \end{vmatrix} \]
By performing determinant calculations, we obtain \( W(t) = 16e^{4t} \). The fact that the Wronskian is not zero confirms that the solutions \( y_1, y_2, y_3 \) are linearly independent at all times \( t \). This is pivotal in assuring that these solutions form a fundamental set for the differential equation.

An initial value problem (IVP) involves finding a specific solution to a differential equation that also meets given initial conditions. In our exercise, we are provided with the conditions:

• \( \phi(0) = 0 \)
• \( \phi'(0) = 1 \)
• \( \phi''(0) = 0 \)

To find the particular solution \( \phi(t) \), we express it as a linear combination of the linearly independent solutions found earlier, in the form \( \phi(t) = C_1 + C_2 e^{2t} + C_3 e^{-2t} \).
Applying the initial conditions involves substituting \( t = 0 \) into the equation and its derivatives. These calculations create a system of equations that allow us to solve for the constants \( C_1, C_2, \) and \( C_3 \).
Solving the system, we find that \( C_1 = 0 \), \( C_2 = \frac{1}{4} \), and \( C_3 = -\frac{1}{4} \). Hence, the particular solution meeting all initial conditions is \( \phi(t) = \frac{1}{4} e^{2t} - \frac{1}{4} e^{-2t} \). This approach ensures that the solution not only satisfies the differential equation but also aligns with the specific starting conditions provided.