91Ó°ÊÓ

The characteristic equation is a fundamental concept when solving linear differential equations with constant coefficients. It is derived from the original differential equation by replacing the derivatives with powers of a variable, traditionally denoted as \( r \). This process transforms a differential equation into a polynomial equation of \( r \), known as the characteristic equation.

For our exercise, the given second-order differential equation is:

• \( 6 y'' - y' - y = 0 \)

By substituting \( y' \) with \( r \) and \( y'' \) with \( r^2 \), we obtain:

• \( 6r^2 - r - 1 = 0 \)

The solution to this polynomial will provide us with the roots (\( r \)) that are critical in constructing the general solution of the differential equation. Essentially, the characteristic equation converts a differential problem into an algebraic one that is often more straightforward to solve.

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The quadratic formula enables us to find the roots of such equations and is given by:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our characteristic equation \( 6r^2 - r - 1 = 0 \), we identify the coefficients: \( a = 6 \), \( b = -1 \), and \( c = -1 \).

Substituting these values into the quadratic formula gives:

• \( r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \)
• \( r = \frac{1 \pm \sqrt{1 + 24}}{12} \)
• \( r = \frac{1 \pm 5}{12} \)

Thus, we obtain the roots for our characteristic equation, which are \( r_1 = \frac{1}{2} \) and \( r_2 = -\frac{1}{3} \). These roots are essential as they guide the formulation of the differential equation's solution.

The general solution of a differential equation like ours provides a formula that encompasses all possible solutions. In the context of a second-order linear homogeneous differential equation with constant coefficients, where the characteristic equation yields two distinct real roots, the general solution is expressed mathematically as:

• \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)

Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions, \( r_1 \) and \( r_2 \) are the roots of the characteristic equation.

Using the roots obtained from the characteristic equation solution \( r_1 = \frac{1}{2} \) and \( r_2 = -\frac{1}{3} \), the general solution becomes:

• \( y(t) = C_1 e^{\frac{1}{2} t} + C_2 e^{-\frac{1}{3} t} \)

This solution describes how the system behaves over time and can be tailored by specifying \( C_1 \) and \( C_2 \) based on additional conditions or constraints, such as initial values.