Chapter 3: Problem 47
In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}+3 y^{\prime}+2 y=4 e^{x} $$
Short Answer
Expert verified
The particular solution of the differential equation is \( y_p = \frac{4}{3} e^{x} \).
Step by step solution
01
Identify the Differential Equation
We are given the second-order linear differential equation with constant coefficients: \[ y'' + 3y' + 2y = 4e^x. \] Our goal is to find a particular solution using the method of variation of parameters.
02
Solve the Homogeneous Equation
The associated homogeneous equation is \[ y'' + 3y' + 2y = 0. \]The characteristic equation is \[ r^2 + 3r + 2 = 0. \] Factoring, we find \( (r + 1)(r + 2) = 0 \), leading to roots \( r = -1 \) and \( r = -2 \). The general solution of the homogeneous equation is \[ y_h = c_1 e^{-x} + c_2 e^{-2x}. \]
03
Identify the Form of the Particular Solution
We are going to use variation of parameters to find a particular solution, \( y_p \), of the non-homogeneous equation. We consider \( y_p = u_1 y_1 + u_2 y_2 \), where \( u_1 \) and \( u_2 \) are functions to be determined, and \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) are solutions of the homogeneous equation.
04
Derive The Equations for u_1 and u_2
For variation of parameters, we use the formulas\[ u_1' y_1 + u_2' y_2 = 0 \] and \[ u_1' y_1' + u_2' y_2' = g(x), \] where \( g(x) = 4e^x. \)This leads us to solving \[\begin{aligned}&u_1' e^{-x} + u_2' e^{-2x} = 0 \&-u_1' e^{-x} - 2u_2' e^{-2x} = 4e^x.\end{aligned}\]
05
Solve for u_1' and u_2'
Use the method of elimination to solve the system. Multiply the first equation by 2 and add to the second equation:\[\begin{aligned}-u_1' e^{-x} - 2u_2' e^{-2x} &= 4e^x, \2u_1' e^{-x} + 2u_2' e^{-2x} &= 0.\end{aligned}\]Add these equations: \[ u_1' e^{-x} = 4e^x \] giving \( u_1' = 4e^{2x}. \)Substitute back to find \( u_2' \): \[ u_2' e^{-2x} = -2e^x - 2 \times 4e^{2x} e^{-2x} = -2e^x. \] Then, \( u_2' = -2e^{3x}. \)
06
Integrate to Find u_1 and u_2
Integrate to find \( u_1 \) and \( u_2 \):\[\begin{aligned}u_1 &= \int 4e^{2x} \, dx = 2e^{2x} + C_1, \u_2 &= \int -2e^{3x} \, dx = -\frac{2}{3}e^{3x} + C_2.\end{aligned}\] We focus on finding a particular solution, so we can choose \( C_1 = C_2 = 0 \).
07
Write the Particular Solution
Substitute back into the particular solution form \[ y_p = u_1 y_1 + u_2 y_2, \] where \[\begin{aligned}y_p &= (2e^{2x}) e^{-x} + \left(-\frac{2}{3}e^{3x}\right) e^{-2x} \&= 2e^{x} - \frac{2}{3}e^{x} \&= \frac{4}{3} e^{x}.\end{aligned}\] This is the particular solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They are crucial in understanding how different variables interact with each other through rates of change. The given problem involves a second-order linear differential equation:
\[ y'' + 3y' + 2y = 4e^x. \]
In this equation, the terms \( y'' \), \( y' \), and \( y \) represent the second derivative, first derivative, and the original function of \( y \) respectively. These equations are called **linear** because they involve linear combinations of the function and its derivatives. When other factors, like the right-hand side, are non-zero, they become non-homogeneous equations, which need special techniques for their solutions, such as the variation of parameters.
\[ y'' + 3y' + 2y = 4e^x. \]
In this equation, the terms \( y'' \), \( y' \), and \( y \) represent the second derivative, first derivative, and the original function of \( y \) respectively. These equations are called **linear** because they involve linear combinations of the function and its derivatives. When other factors, like the right-hand side, are non-zero, they become non-homogeneous equations, which need special techniques for their solutions, such as the variation of parameters.
Homogeneous Equation
A homogeneous equation is a type of differential equation where every term depends on the function or its derivatives. There are no standalone constants or functions that aren't a part of the function's derivatives.
The homogeneous form associated with our given equation is:
\[ y'' + 3y' + 2y = 0. \]
When solving a homogeneous equation, our goal is to find the general solution, often expressed in terms of arbitrary constants. This solution lays the foundation for finding a particular solution to the non-homogeneous equation.For this homogeneous equation, we identify that:
The homogeneous form associated with our given equation is:
\[ y'' + 3y' + 2y = 0. \]
When solving a homogeneous equation, our goal is to find the general solution, often expressed in terms of arbitrary constants. This solution lays the foundation for finding a particular solution to the non-homogeneous equation.For this homogeneous equation, we identify that:
- It uses the characteristic equation to find possible solutions.
- It guides us to write the general solution using exponential functions related to its roots.
Particular Solution
A particular solution is a solution to a non-homogeneous differential equation that accounts for the non-zero right-hand side of the equation.
Using the method of variation of parameters, we approach finding a particular solution for the problem. Here, we attempt to find functions \( u_1 \) and \( u_2 \) such that the particular solution can be expressed as:
\[ y_p = u_1 y_1 + u_2 y_2, \]where \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) are solutions to the homogeneous equation.
The aim is to solve for \( u_1 \) and \( u_2 \) by integrating and determining that they are functions responsible for altering solutions of the homogeneous equation to fit the non-homogeneous form. The calculated particular solution reflects the system's response exclusively to external inputs or disturbances.
Using the method of variation of parameters, we approach finding a particular solution for the problem. Here, we attempt to find functions \( u_1 \) and \( u_2 \) such that the particular solution can be expressed as:
\[ y_p = u_1 y_1 + u_2 y_2, \]where \( y_1 = e^{-x} \) and \( y_2 = e^{-2x} \) are solutions to the homogeneous equation.
The aim is to solve for \( u_1 \) and \( u_2 \) by integrating and determining that they are functions responsible for altering solutions of the homogeneous equation to fit the non-homogeneous form. The calculated particular solution reflects the system's response exclusively to external inputs or disturbances.
Characteristic Equation
The characteristic equation is a critical component in solving homogeneous differential equations. It provides an algebraic method to find solutions of differential equations by turning a differential equation into a solvable polynomial. For our homogenous equation, the characteristic equation is given as:
\[ r^2 + 3r + 2 = 0. \]The roots of this characteristic equation, found through techniques such as factoring, dictate the form of the general solution. In our case, the roots are \( r = -1 \) and \( r = -2 \), simplifying the task of constructing the solution:
\[ r^2 + 3r + 2 = 0. \]The roots of this characteristic equation, found through techniques such as factoring, dictate the form of the general solution. In our case, the roots are \( r = -1 \) and \( r = -2 \), simplifying the task of constructing the solution:
- Root \( r = -1 \) gives rise to the function \( e^{-x} \).
- Root \( r = -2 \) results in \( e^{-2x} \).