Problem 67 - 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. $$ 3.15 y^{(4)}-5.34 y^{\prime \prime}+6.33 y^{\prime}-2.03 y=0 $$

Step-by-Step Solution

Answer

Find the roots of the auxiliary equation to derive the general solution.

Step 1: Identify the Differential Equation Order

The given differential equation \(3.15 y^{(4)} - 5.34 y'' + 6.33 y' - 2.03 y = 0\) is a fourth-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Auxiliary Equation

For a differential equation of the form \(a y^{(n)} + b y^{(n-1)} + \, ... \, + k y = 0\), the auxiliary equation is obtained by replacing derivatives with powers of a variable, typically denoted by \(r\). Here, the auxiliary equation is \(3.15 r^4 - 5.34 r^2 + 6.33 r - 2.03 = 0\).

Step 3: Solve the Auxiliary Equation

Using a computer algebra system (CAS), solve the polynomial \(3.15 r^4 - 5.34 r^2 + 6.33 r - 2.03 = 0\) for \(r\). The solutions could be real or complex, and may repeat.

Step 4: Find the General Solution

The roots found in the previous step are used to write the general solution of the differential equation. For each distinct real root \(r_i\), the solution includes \(C_i e^{r_i t}\). For complex roots \(\alpha \pm \beta i\), the solution includes terms \(e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))\).

Step 5: Simplify the General Solution

Combine the terms in the general solution into a simplified expression. Ensure that all terms are in terms of real functions.

Key Concepts

Auxiliary Equation

The auxiliary equation is a crucial step in solving linear differential equations, especially with constant coefficients. It converts the differential equation into an algebraic problem, making it easier to handle.
For the differential equation provided, the auxiliary equation is formed by replacing each derivative term

• Use the power of a variable, often denoted as \( r \), replacing each derivative of \( y \).
• This turns a differential equation into a polynomial equation.

In the problem provided: - The differential equation is \( 3.15 y^{(4)} - 5.34 y'' + 6.33 y' - 2.03 y = 0 \).- This becomes the polynomial \( 3.15 r^4 - 5.34 r^2 + 6.33 r - 2.03 = 0 \).This transformation is particularly helpful because solving polynomial equations for their roots is a standard procedure, often aided by computer software. Identifying these roots is vital to constructing the general solution of the differential equation.

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is distinguished by its structure and characteristics, which makes it an important type of equation in differential calculus. In our context:

• "Linear" means that each term is either a derivative of \( y \), or \( y \) itself, multiplied by a function of \( t \), but cannot be a product of derivatives or powers of \( y \).
• "Homogeneous" indicates that there is no constant term or function of \( t \) on the right side of the equation, which is instead set to zero.

For example, in the equation provided:- All terms involve derivatives or \( y \), multiplied by constant coefficients.- The right side of the equation is zero, further confirming its homogeneity.This structure allows the equation to be solved using the auxiliary equation method, leading to finding the general solution by interpreting the nature of the roots.

Constant Coefficients

Constant coefficients in a differential equation like the one given are coefficients that do not depend on the independent variable, typically \( t \).
They are significant because they simplify both the formation of the auxiliary equation and the solution process:

• They make the method of writing an auxiliary equation applicable, allowing the problem to become a polynomial equation.
• Additionally, constant coefficients enable easy manipulation using typical algebraic techniques.

In our exercise:- The coefficients \(3.15, -5.34, 6.33, -2.03\) do not vary with \( t \).- Solving the auxiliary equation becomes straightforward as we deal with a polynomial equation, \solving for \( r \) directly impacts the formation of the general solution.Understanding the nature of the coefficients and how they impact the equation aids in the overall solution process.

General Solution

The general solution of a linear homogeneous differential equation with constant coefficients is derived from the roots of the auxiliary equation. Here's how it works:

• For each distinct real root \( r_i \) of the auxiliary equation, the solution has a term of the form \( C_i e^{r_i t} \).
• If there are complex roots, of the form \( \alpha \pm \beta i \), the solution incorporates terms like \( e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \).
• If a root repeats \( k \) times, then polynomial terms such as \( t^k \) multiply each exponential term.

In our example, the specific roots of the polynomial \( 3.15 r^4 - 5.34 r^2 + 6.33 r - 2.03 = 0 \) determine the form and terms of the solution. The resulting general solution encapsulates all possible behaviors of the differential equation, depending on initial conditions or additional constraints that might be applied in specific problems.