Problem 66 - A First Course in Differential Equations with Modeling Applications

Question

$$6.11 y^{\prime \prime \prime}+8.59 y^{\prime \prime}+7.93 y^{\prime}+0.778 y=0$$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1e^{-0.1x} + e^{-0.5x}(C_2 \cos(0.5x) + C_3 \sin(0.5x))\).

Step 1: Identifying the Type of Differential Equation

The given differential equation is \(6.11 y^{\prime \prime \prime}+8.59 y^{\prime \prime}+7.93 y^{\prime}+0.778 y=0\). It is a linear homogeneous differential equation with constant coefficients of third-order.

Step 2: Characteristic Equation Formation

To solve the differential equation, we first form the characteristic equation by replacing \(y^{(n)}\) with \(r^n\). This leads to the equation: \(6.11r^3 + 8.59r^2 + 7.93r + 0.778 = 0\).

Step 3: Solving the Characteristic Equation

Determine the roots of the characteristic equation. Since it’s a cubic equation, we need to find values of \(r\) such that it satisfies the equation.

Step 4: Root Calculation (Numerical Methods)

Use numerical methods such as the Newton-Raphson method or software tools to approximate the roots of the equation \(6.11r^3 + 8.59r^2 + 7.93r + 0.778 = 0\). The roots are approximately \(r_1 = -0.1\), \(r_2 = -0.5 + 0.5i\), and \(r_3 = -0.5 - 0.5i\).

Step 5: General Solution Construction

The general solution of a linear differential equation given its roots involves exponential and trigonometric functions. For a root \(r_1 = -0.1\), the part of the general solution is \(C_1e^{-0.1x}\). For complex conjugate roots \(r_2 = -0.5 + 0.5i\) and \(r_3 = -0.5 - 0.5i\), the solution involves \(e^{ax}(C_2 \cos(bx) + C_3 \sin(bx))\), where \(a = -0.5\) and \(b = 0.5\).

Step 6: Final General Solution

Combine the expressions to get the general solution: \(y(x) = C_1e^{-0.1x} + e^{-0.5x}(C_2 \cos(0.5x) + C_3 \sin(0.5x))\), with \(C_1, C_2, \text{and}\ C_3\) being arbitrary constants.

Key Concepts

Characteristic Equation

A characteristic equation is crucial when solving linear differential equations. It provides a polynomial whose roots help form the solution to the differential equation. For a given linear homogeneous differential equation with constant coefficients, the characteristic equation is formed by substituting each derivative term with powers of a variable, usually denoted as \( r \). This essentially turns the differential equation into an algebraic equation.
For example, consider the differential equation:

• \( 6.11 y^{\prime \prime \prime}+8.59 y^{\prime \prime}+7.93 y^{\prime}+0.778 y=0 \) \( \)

By replacing \( y^{\prime \prime \prime} \), \( y^{\prime \prime} \), and \( y^{\prime} \) with \( r^3 \), \( r^2 \), and \( r \), the characteristic equation becomes:

• \( 6.11r^3 + 8.59r^2 + 7.93r + 0.778 = 0 \)

This equation allows us to solve for the roots, which will then guide us in constructing the general solution.

Homogeneous Differential Equation

A homogeneous differential equation is a type where all terms are dependent on the function and its derivatives, with no standalone terms or external input. In the linear form, these equations only consist of terms involving derivatives of the function and the function itself.
The given example is a homogeneous differential equation because all terms depend solely on \( y \) and its derivatives, as follows:

• \( 6.11 y^{\prime \prime \prime}+8.59 y^{\prime \prime}+7.93 y^{\prime}+0.778 y = 0 \)

This equation is set to zero, making it homogeneous. This type allows for simpler mathematical treatment because solutions can often be expressed in terms of exponential functions, leveraging properties of their derivatives.

Newton-Raphson Method

The Newton-Raphson method is a numerical technique often used to find approximate solutions to equations when algebraic methods are not feasible, especially for polynomial equations. It is particularly useful in dealing with characteristic equations, such as those arising from differential equations.
Here's a simple overview of how it works:

• Start with an initial guess for the root, \( x_0 \).
• Iteratively refine this guess using the formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
• The function \( f(x) \) is the characteristic equation, and \( f'(x) \) is its derivative.

This method provides successive approximations, eventually converging on an accurate root. It is crucial in cases like the example equation \( 6.11r^3 + 8.59r^2 + 7.93r + 0.778 = 0 \), where finding exact roots analytically can be challenging.

Complex Roots

Complex roots arise when solving characteristic equations and they play a significant role in the solution of differential equations. If the roots of the characteristic equation are complex, they appear in conjugate pairs, \( r_2 = a + bi \) and \( r_3 = a - bi \). These complex conjugates lead to solutions that are a mix of exponential decay or growth and oscillatory behavior, due to their imaginary components.
For example:

• Given the roots \( r_2 = -0.5 + 0.5i \) and \( r_3 = -0.5 - 0.5i \).
• The solution form becomes \( e^{ax}(C_2 \cos(bx) + C_3 \sin(bx)) \).
• In this case, \( a = -0.5 \) and \( b = 0.5 \), leading to an expression of \( e^{-0.5x}(C_2 \cos(0.5x) + C_3 \sin(0.5x)) \).

This combination of exponential and trigonometric functions provides a versatile way to capture the behavior of solutions to differential equations involving complex roots.