Problem 52 - A First Course in Differential Equations with Modeling Applications

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$y=c_{1} e^{10 x}+c_{2} x e^{10 x}$$

Step-by-Step Solution

Answer

The differential equation is \( y'' - 20y' + 100y = 0 \).

Step 1: Identify the Solution Form

The given general solution is \( y = c_{1} e^{10x} + c_{2} x e^{10x} \). It consists of terms with exponential functions related to \( e^{10x} \).

Step 2: Identify the Roots

The roots related to the form \( e^{10x} \) and \( x e^{10x} \) are derived from the term \( e^{10x} \). The presence of the term \( x e^{10x} \) indicates a repeated root. Therefore, \( r = 10 \) is a root with multiplicity 2.

Step 3: Write the Characteristic Equation

The characteristic equation corresponding to repeated roots \( r = 10 \) with multiplicity 2 is \( (r - 10)^2 = 0 \).

Step 4: Form the Differential Equation

Translate the characteristic equation to a differential equation with constant coefficients. The operator form is \((D - 10)^2 y = 0\), where \( D \) represents differentiation with respect to \( x \).

Step 5: Expand to Obtain the Differential Equation

Expand \((D - 10)^2\) to find the differential equation: \((D - 10)^2 = D^2 - 20D + 100\). Hence, the differential equation is \( D^2 y - 20Dy + 100y = 0 \), or \( y'' - 20y' + 100y = 0 \).

Key Concepts

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a special type of differential equation where every term is either dependent on the function or its derivatives, and the equation is set to zero. This means that any solutions to the equation are typically combinations of basic solutions that satisfy the equation.
These equations have some distinct properties:

• They always equate to zero: The formula looks like \(a_n y^{(n)} + a_{n-1} y^{(n-1)} + \ldots + a_1 y' + a_0 y = 0\)
• The coefficients \(a_n, a_{n-1}, \ldots, a_1, a_0\) are constant numbers.

Since they equate to zero, homogeneous differential equations model systems where changes in one part are perfectly balanced by changes in another, maintaining an overall constant outcome. For instance, they might represent structures in equilibrium or electrical circuits where the input and output energy balance.

Characteristic Equation

The characteristic equation is a vital tool when solving linear differential equations, especially when dealing with constant coefficients. For a homogeneous linear differential equation of order \(n\), the characteristic equation is derived by assuming solutions of the form \(y = e^{rx}\) and substituting this back into the differential equation.
For example, if the highest derivative in the differential equation is \(y''\), the characteristic equation will be a polynomial of the form \(ar^2 + br + c = 0\) where \(r\) represents the possible roots.
Characteristic equations are essential because:

• The roots of the characteristic equation denote the types of solutions (e.g., real roots, complex roots).
• Multiple roots result in solutions that include polynomial terms (e.g., \(x\), \(x^2\)) as seen in repeated roots.

These make it a critical step in determining the form of the general solution.

Constant Coefficients

Differential equations with constant coefficients have coefficients that remain unchanged with respect to the variable. This simplification allows us to apply methods like the characteristic equation more easily.
Here's why constant coefficients are helpful:

• The differential equation becomes easier to handle mathematically because they don't change.
• We can use exponential functions \(e^{rx}\) as potential solutions.
• They enable the use of algebraic methods to find particular solutions.

The predictability and simplicity of constant coefficients make them a preferred choice when modeling physical systems, like mechanical vibrations or electrical circuits, where material properties (coefficients) don't change over time.

General Solution

The general solution of a differential equation encompasses all possible specific solutions. By finding the general solution, you essentially have the full set of answers for any initial conditions or specific scenarios.
Generally, the steps involve:

• Finding the roots of the characteristic equation (these determine terms in your solution).
• Constructing solutions based on the type of roots (real, repeated, complex).

For instance, if a root \(r = 10\) has multiplicity 2, the general solution might look like \(y = c_1 e^{10x} + c_2 x e^{10x}\). Here, \(c_1\) and \(c_2\) are arbitrary constants determined by initial conditions.
Thus, while the characteristic equation gives us the form of solutions, the general solution provides the actual equations needed to describe every possible state of the system.