Chapter 3: Problem 9
Find a particular solution \(y_{p}\) of the given equation. In all these problems, primes denote derivatives with respect to \(x\). $$ y^{\prime \prime}+2 y^{\prime}-3 y=1+x e^{x} $$
Short Answer
Expert verified
The particular solution is \( y_p = -\frac{1}{3} + \frac{1}{2}xe^x \).
Step by step solution
01
Identify the Non-homogeneous Term
The differential equation is \( y'' + 2y' - 3y = 1 + xe^x \). The non-homogeneous term on the right-hand side is \( 1 + xe^x \). This suggests that a particular solution \( y_p \) could be the sum of particular solutions for \( 1 \) and \( xe^x \) separately.
02
Solve for the Particular Solution of Constant Term
To find a particular solution for the constant \( 1 \), we assume \( y_{p1} = A \), where \( A \) is a constant. Substituting into the differential equation, \( 0 + 0 - 3A = 1 \), we get \( A = -\frac{1}{3} \). Thus, one particular solution is \( y_{p1} = -\frac{1}{3} \).
03
Solve for the Particular Solution of Non-constant Term
For the term \( xe^x \), assume a particular solution of the form \( y_{p2} = (Bx + C)e^x \), as the exponential suggests such a form. Knowing the form, calculate its derivatives: \( y'_{p2} = (Bx + B + C)e^x \) and \( y''_{p2} = (Bx + 2B + C)e^x \). Substituting into the differential equation:
04
Substitute and Solve for Coefficients
Substitute \( y_{p2} \), \( y'_{p2} \), and \( y''_{p2} \) into the LHS of the original equation: \[ (Bx + 2B + C)e^x + 2(Bx + B + C)e^x - 3(Bx + C)e^x = xe^x \] Simplifying, we get an equation in terms of coefficients: \( 2Bxe^x + (4B)e^x = xe^x \). Equating coefficients, \( 2B = 1 \) gives \( B = \frac{1}{2} \) and \( 4B = 0 \). Checking, \( C=0 \). Thus, \( y_{p2} = \frac{1}{2}xe^x \).
05
Combine Particular Solutions
The overall particular solution \( y_p \) is the sum of the solutions found: \( y_p = y_{p1} + y_{p2} = -\frac{1}{3} + \frac{1}{2}xe^x \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Non-homogeneous Differential Equations
A non-homogeneous differential equation is a type of differential equation that includes a non-zero term on one side known as the non-homogeneous term, or the forcing function. In these equations, the left side typically involves derivatives of a function, while the right side is a function of the independent variable, such as in this example, the equation \[ y'' + 2y' - 3y = 1 + xe^x \]. This equation is non-homogeneous due to the presence of the term \( 1 + xe^x \), which is not equal to zero.
Unlike homogeneous differential equations, where the solution can be expressed as a linear combination of basic solutions due to the right-hand side being zero, non-homogeneous equations require us to find both a complementary solution and a particular solution.
Understanding the differences is essential:
Unlike homogeneous differential equations, where the solution can be expressed as a linear combination of basic solutions due to the right-hand side being zero, non-homogeneous equations require us to find both a complementary solution and a particular solution.
Understanding the differences is essential:
- Homogeneous Equations: All terms are purely in terms of the function and its derivatives.
- Non-homogeneous Equations: Involve additional terms independent from the solution’s function.
Finding a Particular Solution
The particular solution of a non-homogeneous differential equation is a specific solution that addresses the non-homogeneous part of the equation. For the given exercise, the solution comprises finding individual particular solutions to each component of the non-homogeneous term \(1 + xe^x\), and then combining them.
You can use these steps to find a particular solution:
You can use these steps to find a particular solution:
- Identify different parts of the non-homogeneous term. Here, they are the constant 1 and the term \(xe^x\).
- Assume a form for the particular solution tailored to each part. For example, assume a constant form \(y_{p1} = A\) for 1, and a form including the similar structures \((Bx + C)e^x\) for \(xe^x\).
- Determine the constants by substituting back into the equation, ensuring the left-hand side equates to the exact non-homogeneous term on the right.
Using the Method of Undetermined Coefficients
The method of undetermined coefficients is a popular technique for finding particular solutions of linear non-homogeneous differential equations, especially when the non-homogeneous term has special forms like polynomials, exponentials, or trigonometric functions.
This method involves assuming a form for the particular solution that mimics the non-homogeneous term. Coefficients are initially undetermined and hence the name 'undetermined coefficients'. Here’s how it works:
This method involves assuming a form for the particular solution that mimics the non-homogeneous term. Coefficients are initially undetermined and hence the name 'undetermined coefficients'. Here’s how it works:
- Make an educated guess about the form of the particular solution. This guess is often based on the type of the non-homogeneous term.
- In case of duplicates with the complementary solution, multiply the guess with powers of \(x\) until distinct.
- Substitute your guess into the differential equation to find relationships between the coefficients by equating terms.