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The characteristic equation is a critical concept in solving linear differential equations. It essentially transforms a differential equation into an algebraic equation in terms of a new variable, typically denoted as \(r\). This is extremely useful, as algebraic equations are often simpler to solve than differential equations. To derive the characteristic equation from a linear differential equation like \(8y'' + 14y' - 15y = 0\), we substitute each derivative with powers of \(r\):

• \( y'' \) becomes \(r^2\)
• \( y' \) becomes \(r\)
• \( y \) remains as \(1\)

This substitution transforms the differential equation into a quadratic equation: \(8r^2 + 14r - 15 = 0\). By solving this characteristic equation, we find the roots \(r\) that are essential for determining the form of the solution to the differential equation. This step is foundational, as it sets the stage for using additional mathematical tools like the quadratic formula.

The quadratic formula is a tool used to find the solutions of quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). In the case of our characteristic equation \(8r^2 + 14r - 15 = 0\), the quadratic formula helps us determine the values of \(r\) by using the coefficients \(a\), \(b\), and \(c\).The quadratic formula is given by:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here's how it applies:

• For \(a = 8\), \(b = 14\), and \(c = -15\), first compute the discriminant: \(b^2 - 4ac\).
• The discriminant is \(14^2 - 4 \times 8 \times (-15) = 676\). Since the discriminant is positive, it indicates there are two distinct real roots.
• Using these values in the quadratic formula results in the solutions \(r_1 = \frac{3}{4}\) and \(r_2 = -\frac{5}{2}\).

These solutions \(r_1\) and \(r_2\) are crucial for constructing the general solution for the differential equation.

The general solution of a linear differential equation with constant coefficients is formulated once the roots of the characteristic equation are found. In our example, the roots \(r_1 = \frac{3}{4}\) and \(r_2 = -\frac{5}{2}\) emerged from the characteristic equation.Given that these roots are distinct and real, the general solution takes the form of a linear combination of exponentials:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] Substituting the specific roots, the general solution becomes:\[ y(t) = C_1 e^{\frac{3}{4} t} + C_2 e^{-\frac{5}{2} t} \]Here:

• \(C_1\) and \(C_2\) are constants that can be determined if initial conditions are provided. These constants essentially adjust the solution to match specific starting conditions of the problem.
• The terms \(e^{r_1 t}\) and \(e^{r_2 t}\) reflect the exponential nature of solutions to linear differential equations.

Understanding the general solution is paramount, as it gives a complete representation of all possible solutions the differential equation can have under given conditions. It illustrates how the behavior of the system described by the differential equation evolves over time.