Problem 68 - A First Course in Differential Equations with Modeling Applications

Question

$$y^{(4)}+2 y^{\prime \prime}-y^{\prime}+2 y=0$$

Step-by-Step Solution

Answer

The general solution is \(y(t) = C_1 e^t + C_2 e^{-t} + C_3 \cos(t) + C_4 \sin(t)\).

Step 1: Define the Characteristic Equation

Since the differential equation is linear and has constant coefficients, we first form the characteristic equation by assuming a solution of the form \(y = e^{rt}\). This gives us \(r^4 + 2r^2 - r + 2 = 0\).

Step 2: Simplify the Characteristic Equation

Attempt to simplify or factor the characteristic equation \(r^4 + 2r^2 - r + 2 = 0\). Look for factorable components or apply known methods like synthetic division or trial roots to find one root of the polynomial. Unfortunately, this polynomial does not simplify easily, suggesting synthetic or numerical methods for real or complex roots.

Step 3: Find Real/Complex Roots

Using numerical methods or trial and error, we find the roots of the equation are \(r = 1\), \(r = -1\), and \(r = \pm i\), which can be verified by substituting back into the characteristic equation.

Step 4: Construct the General Solution

Using the roots found, the general solution of the differential equation is a combination of exponential and sinusoidal functions: \[y(t) = C_1 e^t + C_2 e^{-t} + C_3 \cos(t) + C_4 \sin(t)\]Where \(C_1, C_2, C_3, \) and \(C_4\) are arbitrary constants determined by initial conditions.

Key Concepts

Characteristic Equation

When solving differential equations, especially linear ones with constant coefficients, finding the characteristic equation is the very first step. This equation is derived from assuming that the solution has the form of an exponential function, specifically, \( y = e^{rt} \). By substituting this assumed solution into the differential equation, you replace the derivatives with powers of \( r \). The resulting algebraic equation for \( r \) is the characteristic equation.

• It serves as an essential tool for determining the roots that will form part of the solution to the differential equation.
• In the given problem, the characteristic equation is \( r^4 + 2r^2 - r + 2 = 0 \).
• This equation encapsulates the essence of the differential equation by transforming a differential problem into an algebraic one.

Understanding the characteristic equation is crucial as it forms the bridge between the differential equation and its solution through its roots.

General Solution

The general solution to a differential equation comes from the roots of its characteristic equation. Once the roots are found, they inform us about the form of the solution. The general solution is typically a mix of exponential, sinusoidal, or polynomial terms, depending on whether the roots are real, complex, or repeated.

• For each distinct real root, you have an exponential term \( C e^{rt} \).
• For each pair of complex conjugate roots, \( a \pm bi \), the solution involves sinusoidal functions, \( e^{at}(C_1 \cos(bt) + C_2 \sin(bt)) \).

In the example, the general solution formed is: \[y(t) = C_1 e^t + C_2 e^{-t} + C_3 \cos(t) + C_4 \sin(t)\]Each term corresponds to a root of the characteristic equation, showing how the algebraic properties of the roots influence the type of functions in the solution.

Complex Roots

Complex roots often appear in characteristic equations and are critical in defining a part of the general solution. If the characteristic equation has complex roots, they typically come in conjugate pairs \( a \pm bi \).

• These complex roots result in a general solution involving trigonometric functions.
• This is due to Euler’s formula, which relates complex exponentials to sine and cosine functions: \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \).

In the given exercise, the complex roots are \( \pm i \). Because of these complex roots, we derive terms like \( \cos(t) \) and \( \sin(t) \) in the solution, representing oscillatory behavior. Such terms reflect phenomena that repeat over time, a fundamental concept across physics and engineering.

Exponential and Sinusoidal Functions

The presence of exponential and sinusoidal functions in solutions of differential equations is integral to depicting various types of behaviors, especially periodic and transient responses.

• Exponential functions \( e^{at} \) tend to describe growth or decay, depending on whether \( a \) is positive or negative.
• Sinusoidal functions, derived from complex roots, represent periodic, oscillatory behavior, echoing the natural cycles found in physical systems.

For the given differential equation, the solution combines these functions to resemble a full picture of potential system behaviors:

• \( e^t \) and \( e^{-t} \) imply growth and decay respectively.
• \( \cos(t) \) and \( \sin(t) \) show periodicity.

Together, these functions allow for the accurate modeling of systems such as electrical circuits, pendulums, and even economic cycles in real-world applications.