91Ó°ÊÓ

A differential equation often seems complex because it involves derivatives. To make it simpler, particularly for a third-order differential equation like the one given, we use the characteristic equation. This equation transforms the problem from one involving derivatives to polynomial equations, which are easier to handle. Imagine derivatives become terms like powers in algebra, and each derivative, such as \( y'\) or \( y''\), is replaced by powers of \( r \). For the differential equation:
\[ y^{\prime \prime \prime}-2 y^{\prime \prime}-3 y^{\prime}=0 \]
We substitute to get the characteristic equation:
\[ r^3 - 2r^2 - 3r = 0 \]
This mimics the structure of our original equation, turning it into a more straightforward polynomial expression.

Factoring is a crucial skill in both algebra and differential equations. It breaks down a complicated expression into simpler, more manageable factors. For the characteristic equation of this problem, initially expressed as:
\[ r^3 - 2r^2 - 3r = 0 \]
Factoring out common terms is the first step. Here, \( r \) is a common factor, leading to:
\[ r(r^2 - 2r - 3) = 0 \]
Then we focus on factoring the quadratic expression \( r^2 - 2r - 3 \). Finding the pairs that multiply to give the constant term and add to the middle coefficient helps. After careful inspection, we get:
\[ r^2 - 2r - 3 = (r - 3)(r + 1) \]
These multiplication and addition combinations are the key to successfully solving the equation.

Once we have factored the equation, we can use the roots to construct the general solution. The general solution of a third-order differential equation is composed of terms formed from the roots of the characteristic equation:
\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} \]
In this context, \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants that arise because we haven't provided specific initial or boundary conditions. For the roots \( r = 0\), \( r = 3\), and \( r = -1\), the solution becomes:
\[ y(t) = C_1 e^{0t} + C_2 e^{3t} + C_3 e^{-t} \]
Since \( e^{0t} \) simplifies to 1, the final general solution is:
\[ y(t) = C_1 + C_2 e^{3t} + C_3 e^{-t} \]
This compact expression provides the form of all possible solutions to the original differential equation, dependent on the constants determined by specific conditions.

The roots of polynomial equations are vital because they determine the solutions to the corresponding differential equations. In our factored equation:
\[ r(r - 3)(r + 1) = 0 \]
Each factor represents a potential root. When any of these factors equals zero, the entire expression is zero. Thus, the roots are determined by solving:

• \( r = 0 \)
• \( r - 3 = 0 \) giving \( r = 3 \)
• \( r + 1 = 0 \) giving \( r = -1 \)

These roots are straightforward because they are real and distinct. However, for more complex circumstances, roots could be repeated or complex numbers, changing how we formulate the solution of the differential equation. Understanding these roots is essential for constructing any solution.