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The characteristic equation plays a crucial role when solving second-order linear differential equations with constant coefficients. This type of differential equation, like the one provided in the problem (5y'' - y' = 3y), is typically solved by transforming it into a characteristic equation. This involves replacing each derivative with a power of a variable, often denoted as \(r\).Here's how it works:- Replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) itself with 1.- This substitution transforms the given differential equation into a polynomial equation in terms of \(r\).For the given equation, this transformation results in the characteristic equation \(5r^2 - r = 3\). Modifying it into standard quadratic form, we get \(5r^2 - r - 3 = 0\). Solving this characteristic equation, we uncover the roots that are crucial in constructing the general solution for the differential equation.

The quadratic formula is a fundamental tool used to find the roots of quadratic equations. These equations are of the form \(ax^2 + bx + c = 0\). In the context of differential equations, specifically when dealing with the characteristic equation, it becomes essential to determine the roots that will form the solution basis.The quadratic formula is expressed as:\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here:- \(a\), \(b\), and \(c\) are coefficients from the characteristic equation.- The expression under the square root, \(b^2 - 4ac\), is known as the discriminant.In our example, the characteristic equation \(5r^2 - r - 3 = 0\) has the coefficients \(a = 5\), \(b = -1\), and \(c = -3\). Plugging these into the quadratic formula provides:- Discriminant \(b^2 - 4ac = 61\), indicating two distinct real roots since it is positive.- Solving yields \(r_1 = \frac{1 + \sqrt{61}}{10}\) and \(r_2 = \frac{1 - \sqrt{61}}{10}\). These roots are pivotal in formulating the general solution to the differential equation.

Second-order linear differential equations are a type of differential equation where the highest derivative is the second derivative. These equations form the backbone for many physical and engineering applications.In general, they can be portrayed as:\[ ay'' + by' + cy = f(x)\]- When the equation is homogeneous (like the original equation, where \(f(x) = 0\)), solutions often involve finding specific functions that satisfy the equation.The solution process involves:

• Constructing the characteristic equation
• Solving it to find roots
• Using these roots to write the general solution

For equations with constant coefficients, like \(5y'' - y' = 3y\), the characteristic equation simplifies the process of finding solutions. The roots, once determined, furnish a solution expressed in terms of exponentials:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]This format is fundamental in mathematical modeling, representing varied phenomena from vibrations and heat conduction to electrical circuits.