Problem 65 - Advanced Engineering Mathematics

Question

In Problems 65-68, use a computer either as an aid in solving the auxiliary equation or as a means of directly obtaining the general solution of the given differential equation. If you use a CAS to obtain the general solution, simplify the output and, if necessary, write the solution in terms of real functions. $$ y^{\prime \prime \prime}-6 y^{\prime \prime}+2 y^{\prime}+y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^t + C_2 e^{2t} + C_3 e^{3t} \).

Step 1: Understanding the Problem

The given differential equation is \( y''' - 6y'' + 2y' + y = 0 \). We need to solve this third-order linear differential equation using computer software or computer algebra systems (CAS).

Step 2: Using CAS to Find the Characteristic Equation

Enter the differential equation into a CAS to find the characteristic polynomial. For the equation \( y''' - 6y'' + 2y' + y = 0 \), the characteristic polynomial is \( r^3 - 6r^2 + 2r + 1 = 0 \).

Step 3: Solving the Characteristic Polynomial

Use the CAS to solve the characteristic polynomial equation \( r^3 - 6r^2 + 2r + 1 = 0 \). This yields the roots: \( r_1 = 1 \), \( r_2 = 2 \), and \( r_3 = 3 \).

Step 4: Construct the General Solution from Roots

Since the roots \( r = 1, 2, 3 \) are all real and distinct, the general solution to the differential equation is given by: \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} \).

Step 5: Writing the Solution in Real Functions

Substituting the values of the roots, the general solution becomes: \( y(t) = C_1 e^t + C_2 e^{2t} + C_3 e^{3t} \), where \( C_1, C_2, \) and \( C_3 \) are arbitrary constants.

Key Concepts

Characteristic Equation

In the study of differential equations, the characteristic equation is crucial in finding solutions to linear differential equations. Specifically, in the context of third-order differential equations like the one given, the characteristic equation is a polynomial derived from substituting derivatives with powers of a variable, typically denoted by \( r \).
To form the characteristic equation, each derivative \( y^{(n)} \) in the differential equation is replaced by \( r^n \).
For example, given the equation \( y''' - 6y'' + 2y' + y = 0 \), the characteristic equation is acquired by replacing these derivatives as:

• \( y''' \) becomes \( r^3 \)
• \( y'' \) becomes \( r^2 \)
• \( y' \) becomes \( r \)
• \( y \) becomes \( 1 \)

Therefore, the characteristic equation is \( r^3 - 6r^2 + 2r + 1 = 0 \). Solving this equation helps find the roots, which are essential in constructing the general solution to the differential equation.

Differential Equations Using CAS

Computer Algebra Systems (CAS) are powerful tools used in handling complex computations, especially in solving differential equations. They simplify and expedite the solving process, which might otherwise be cumbersome by hand. When you use a CAS, it acts as an analytical companion, allowing you to focus on understanding the solution process and implications instead of just arithmetic.
Using CAS for our equation begins by inputting the differential equation \( y''' - 6y'' + 2y' + y = 0 \) to directly get the characteristic equation, \( r^3 - 6r^2 + 2r + 1 = 0 \).
After obtaining the characteristic polynomial, the CAS performs complex factorization calculations, quickly providing us with the roots \( r_1 = 1 \), \( r_2 = 2 \), and \( r_3 = 3 \).
By using a CAS, students can verify their hand-done computations, explore different solution methods, and gain deeper insights into the nature of differential equations.

General Solution of Differential Equations

Finding the general solution of a third-order differential equation requires determining the equation's roots from its characteristic polynomial. The nature of these roots—real and distinct, repeated, or complex—affects the form of the solution.
For the given differential equation, the roots were found to be real and distinct: \( r_1 = 1 \), \( r_2 = 2 \), and \( r_3 = 3 \).
This results in the following form for the general solution:

• Each root corresponds to an exponential term \( e^{r_i t} \).
• Thus, the solution structure is \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} \).

Substituting the found roots, we derive: \( y(t) = C_1 e^t + C_2 e^{2t} + C_3 e^{3t} \), where \( C_1, C_2, \) and \( C_3 \) are arbitrary constants determined by initial or boundary conditions, conveying how the system behaves over time.