Problem 56 - Calculus Volume 3

Question

Solve the following equations using the method of undetermined coefficients. $$ y^{\prime \prime}-6 y^{\prime}+5 y=e^{-x} $$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 e^{x} + C_2 e^{5x} + \frac{1}{12} e^{-x} \).

Step 1: Identify the Problem Type

The differential equation given is a second-order linear non-homogeneous differential equation with constant coefficients. The equation is \( y'' - 6y' + 5y = e^{-x} \). The non-homogeneous part is \( e^{-x} \), which indicates the method of undetermined coefficients is applicable.

Step 2: Solve the Homogeneous Equation

First, solve the homogeneous equation \( y'' - 6y' + 5y = 0 \). The characteristic equation is \( r^2 - 6r + 5 = 0 \). Factor this into \((r-1)(r-5)=0\). The roots are \( r = 1 \) and \( r = 5 \). Thus, the general solution to the homogeneous equation is \( y_h = C_1 e^{x} + C_2 e^{5x} \).

Step 3: Formulate a Particular Solution

Since the non-homogeneous term is \( e^{-x} \), we assume a particular solution of the form \( y_p = A e^{-x} \), where \( A \) is a constant to be determined. This form is chosen because \( e^{-x} \) is an exponential function.

Step 4: Substitute and Solve for Coefficients

Substitute \( y_p = A e^{-x} \) into the original differential equation. First, calculate the derivatives: \( y_p' = -A e^{-x} \) and \( y_p'' = A e^{-x} \).Substitute these into the equation:\[ A e^{-x} - 6(-A e^{-x}) + 5A e^{-x} = e^{-x} \]\[ (A + 6A + 5A) e^{-x} = e^{-x} \]\[ 12A e^{-x} = e^{-x} \]Divide by \( e^{-x} \):\[ 12A = 1 \]Thus, \( A = \frac{1}{12} \). The particular solution is \( y_p = \frac{1}{12} e^{-x} \).

Step 5: Write the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions: \( y = y_h + y_p \).Thus, \[ y = C_1 e^{x} + C_2 e^{5x} + \frac{1}{12} e^{-x} \].

Key Concepts

second-order linear differential equation

A second-order linear differential equation involves derivatives up to the second order. The general form of such an equation is \( ay'' + by' + cy = g(x) \), where \( a \), \( b \), and \( c \) are constants, \( y \) is the function of \( x \), and \( g(x) \) is a given function. These equations naturally occur in physics and engineering to describe systems like oscillations or circuits.

The key here is understanding that these equations represent a balance of forces. The goal is often to find the function \( y(x) \) that satisfies the equation.

• The term \( ay'' \) represents the acceleration or concavity of the function.
• The term \( by' \) is related to velocity or slope.
• The term \( cy \) signifies position or the function value itself.

By solving these equations, we can predict how a system behaves over time.

homogeneous equation

A homogeneous differential equation is a special type of differential equation where the right side is zero, e.g., \( y'' - 6y' + 5y = 0 \). Such equations arise when a system has no external forces or inputs acting on it.

To solve it, we find its characteristic equation. In this case, it is \( r^2 - 6r + 5 = 0 \). We solve this quadratic equation to find the roots, which gives us insight into the behavior of the solutions.

The characteristic equation for this homogeneous differential equation factors neatly as \( (r-1)(r-5)=0 \), giving roots \( r = 1 \) and \( r = 5 \).

This results in a general solution of \( y_h = C_1 e^{x} + C_2 e^{5x} \), where \( C_1 \) and \( C_2 \) are constants.

Each term in this solution represents a mode of the system's natural response without external inputs.

particular solution

The particular solution resolves how a system responds to external forces or inputs represented by \( g(x) \) in the original equation. Here, \( g(x) = e^{-x} \), meaning our differential equation is non-homogeneous.

For this, the method of undetermined coefficients is invaluable. We guess a solution form that resembles the non-homogeneous term. Since \( e^{-x} \) matches a simple exponential function, we guess \( y_p = A e^{-x} \).

The idea is to substitute this assumed particular form into the differential equation and solve for \( A \).

After substituting and simplifying, we find \( A = \frac{1}{12} \). This means our particular solution is \( y_p = \frac{1}{12} e^{-x} \).

This specific solution models how the system would behave uniquely under the influence of the term \( e^{-x} \).

characteristic equation

The characteristic equation is fundamental for solving homogeneous linear differential equations. It's derived from trying to solve the homogeneous part by guessing a solution of the form \( e^{rx} \). For the equation \( y'' - 6y' + 5y = 0 \), substituting \( y = e^{rx} \) yields \( ar^2 + br + c = 0 \).

In this exercise, the characteristic equation is \( r^2 - 6r + 5 = 0 \). Solving this quadratic gives the roots \( r = 1 \) and \( r = 5 \), which inform the general solution of the homogeneous equation.

The roots indicate exponential solutions, where each root yields a term of the form \( C e^{rx} \).

• A single real root leads to terms like \( e^{rx} \) in the general solution.
• Distinct real roots indicate separate terms without overlap.
• Complex roots would yield oscillatory solutions, involving sine and cosine functions.

Recognizing the type and number of roots helps form the full solution to the differential equation.