91Ó°ÊÓ

When you have a homogeneous linear differential equation, such as the one given in the exercise, finding the characteristic equation is a crucial step. This process involves transforming the differential equation into a simpler algebraic form.
Consider the differential equation:

• 9y^{\prime\prime\prime} + 0.6y^{\prime\prime} + 0.01y^{\prime} = 0

To move to the characteristic equation, replace derivatives with powers of a new variable, often denoted by \( r \). Specifically, for each derivative replace \( y^{\prime} \) with \( r \cdot y \), \( y^{\prime\prime} \) with \( r^2 \cdot y \), and \( y^{\prime\prime\prime} \) with \( r^3 \cdot y \). This gives you:

• 9r^3 + 0.6r^2 + 0.01r = 0

By finding the characteristic equation, you're setting the stage to determine the roots, which directly relate to the solution of the differential equation.

Solving quadratic equations is a frequent requirement in mathematics. The quadratic formula provides a reliable method for finding the roots of any quadratic equation of the form:

• ax^2 + bx + c = 0

The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In the context of solving the characteristic equation, you need this formula to find roots of the quadratic part of the equation.
For the equation \( 9r^2 + 0.6r + 0.01 = 0 \), we identified:

• a = 9
• b = 0.6
• c = 0.01

Calculating the discriminant \( b^2 - 4ac \) lets you determine the nature of the roots. For this particular equation, the discriminant is \( 0.6^2 - 4 \times 9 \times 0.01 = 0 \). A discriminant of zero means there's a repeated real root.

A homogeneous linear differential equation is one in which every term is a derivative of the unknown function, and the equation equals zero. This type of equation is fundamental in many physical and mathematical applications. In this example:

• 9y^{\prime\prime\prime} + 0.6y^{\prime\prime} + 0.01y^{\prime} = 0

The process of solving these involves assuming solutions in an exponential form, \( y = e^{rt} \), leading to the characteristic equation.
Homogeneous linear differential equations have the form:

• a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y^{\prime} + a_0 y = 0

Solving involves finding the roots of the characteristic equation. The roots determine the type of general solution we propose:

• Simple roots lead to simple exponential solutions.
• Repeated roots introduce polynomial terms multiplied with the exponentials.

Understanding these properties allows us to construct the general solution for the differential equation.