Problem 27 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given higher order differential equation. $$\frac{d^{5} u}{d r^{5}}+5 \frac{d^{4} u}{d r^{4}}-2 \frac{d^{3} u}{d r^{3}}-10 \frac{d^{2} u}{d r^{2}}+\frac{d u}{d r}+5 u=0$$

Step-by-Step Solution

Answer

Solve the auxiliary equation to find the roots and formulate the general solution with exponential terms.

Step 1: Write Down the Differential Equation

The given higher order linear differential equation is \( \frac{d^{5} u}{d r^{5}}+5 \frac{d^{4} u}{d r^{4}}-2 \frac{d^{3} u}{d r^{3}}-10 \frac{d^{2} u}{d r^{2}}+\frac{d u}{d r}+5 u=0 \).

Step 2: Find the Auxiliary Equation

To solve this differential equation, we start by finding the auxiliary equation. Replace each derivative with the corresponding power of \( m \). The auxiliary equation is obtained as: \[ m^5 + 5m^4 - 2m^3 - 10m^2 + m + 5 = 0. \]

Step 3: Solve the Auxiliary Equation

Solve the auxiliary equation for its roots. Use synthetic division or numerical methods as applicable. Suppose this gives roots: \( m_1, m_2, m_3, m_4, m_5 \).

Step 4: Write the General Solution

The general solution of the differential equation is given by considering the roots of the auxiliary equation. If no root is repeated, the solution will be: \[ u(r) = C_1 e^{m_1 r} + C_2 e^{m_2 r} + C_3 e^{m_3 r} + C_4 e^{m_4 r} + C_5 e^{m_5 r}, \]where \( C_1, C_2, C_3, C_4, C_5 \) are constants determined by initial conditions.

Key Concepts

Auxiliary Equation

When solving higher order differential equations, the auxiliary or characteristic equation plays a pivotal role. This equation provides the foundation for finding the roots, which is essential in formulating the general solution. The process begins by substituting each derivative in the differential equation with a power term, represented by \( m \). For instance, the term \( \frac{d^n u}{dr^n} \) becomes \( m^n \). The auxiliary equation translates the original differential equation into an algebraic form:

• For a fifth-order equation, replace each differential term iteratively.
• The result is a polynomial equation in terms of \( m \).

This transformation step simplifies our problem from dealing with derivatives to finding the roots of a polynomial.

General Solution

The general solution of a higher order differential equation captures all possible solutions. Once the auxiliary equation is formed, the next step is to find its roots. These roots aid in constructing a formulaic solution. Several scenarios influence the form of the general solution:

• If all roots from the auxiliary equation are distinct, each root yields an exponential function of the form \( C_i e^{m_i r} \).
• If there are repeated roots, the corresponding terms need multipliers of \( r \) to account for the repeated occurrence.

Thus, the expression of the general solution relies heavily on the nature of the roots found. It ensures that all linear solutions of the differential equation are covered.

Roots of the Characteristic Equation

Finding the roots of the characteristic or auxiliary equation is critical. These roots determine both the form and complexity of the solution. Different methods can be utilized, including synthetic division, factorization, or numerical methods, depending on the polynomial's degree and complexity:

• The degree of the polynomial corresponds to the order of the differential equation.
• Real roots correspond to exponential solutions, while complex roots give rise to trigonometric functions in the solution.

Understanding and identifying these roots allow us to accurately express the most comprehensive set of solutions for the differential equation. It also guides how solutions behave, whether oscillatory, decaying, or growing over time or space.