Chapter 19: Problem 2
Find a particular integral for the equation $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=5 \mathrm{e}^{3 t} $$
Short Answer
Expert verified
The particular integral is \(x_p(t) = \frac{5}{2}e^{3t}\).
Step by step solution
01
Identify the type of non-homogeneous term
Examine the right-hand side of the equation: \(5e^{3t}\). This term is an exponential function, suggesting that the trial solution for a particular integral will also likely be an exponential function.
02
Propose the form of the particular integral
Since the non-homogeneous term is \(5e^{3t}\), propose a trial solution of the form \(x_p(t) = Ae^{3t}\), where \(A\) is a constant to be determined.
03
Differentiate the trial solution
Compute the first and second derivatives of \(x_p(t) = Ae^{3t}\). The first derivative is \(\frac{\mathrm{d} x_p}{\mathrm{d} t} = 3Ae^{3t}\) and the second derivative is \(\frac{\mathrm{d}^2 x_p}{\mathrm{d} t^2} = 9Ae^{3t}\).
04
Substitute into the differential equation
Substitute \(x_p(t) = Ae^{3t}\), \(\frac{\mathrm{d} x_p}{\mathrm{d} t} = 3Ae^{3t}\), and \(\frac{\mathrm{d}^2 x_p}{\mathrm{d} t^2} = 9Ae^{3t}\) into the left-hand side of the differential equation: \[9Ae^{3t} - 3(3Ae^{3t}) + 2Ae^{3t} = 5e^{3t}.\]
05
Simplify and solve for A
Combine like terms: \[9Ae^{3t} - 9Ae^{3t} + 2Ae^{3t} = 5e^{3t},\]which simplifies to \[2Ae^{3t} = 5e^{3t}.\] Cancel \(e^{3t}\) from both sides (assuming \(e^{3t} eq 0\)) and solve for \(A\): \[2A = 5 \Rightarrow A = \frac{5}{2}.\]
06
Write the particular integral
Using the value \(A = \frac{5}{2}\) found in Step 5, the particular integral is \(x_p(t) = \frac{5}{2}e^{3t}\).
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.
Particular Integral
In the context of differential equations, a particular integral is a specific solution to a non-homogeneous differential equation. This integral differs from the complementary solution, which relates only to the homogeneous components of the equation. The process of finding a particular integral involves seeking a function that satisfies the entire non-homogeneous equation.
For the given problem, the equation is \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d}t}+2 x=5 \mathrm{e}^{3 t}\). To find a particular integral, we aim to locate a function that directly addresses the non-homogeneous part, here represented by \(5e^{3t}\). While the complementary solution tackles the equation without a non-homogeneous term, the particular integral requires focusing on the differentials and constants present in the complete form of the equation.
The systematic approach involves:
For the given problem, the equation is \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d}t}+2 x=5 \mathrm{e}^{3 t}\). To find a particular integral, we aim to locate a function that directly addresses the non-homogeneous part, here represented by \(5e^{3t}\). While the complementary solution tackles the equation without a non-homogeneous term, the particular integral requires focusing on the differentials and constants present in the complete form of the equation.
The systematic approach involves:
- Identifying the type of non-homogeneous term.
- Proposing a trial solution that reflects the structure of this term.
- Substituting the trial solution into the differential equation.
- Solving for any unknown constants within the trial solution.
Non-Homogeneous Differential Equation
A non-homogeneous differential equation includes an extra term not present in its homogeneous counterpart. This term is what principally drives the need for a particular integral. In our exercise, the equation is given as \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d}t} + 2x = 5 \mathrm{e}^{3t}\).
The non-homogeneous part, \(5e^{3t}\), sets it apart from the homogeneous equation \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d}t} + 2x = 0\). This type of equation consists of two main sections:
The non-homogeneous part, \(5e^{3t}\), sets it apart from the homogeneous equation \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d}t} + 2x = 0\). This type of equation consists of two main sections:
- The homogeneous solution: This utilizes the characteristic equation derived from setting the differential equation equal to zero.
- The particular solution: Focuses on solving the specific non-homogeneous function, \(5e^{3t}\) in this case.
Exponential Function Trial Solution
The choice of a trial solution is critical in solving for the particular integral in non-homogeneous differential equations. In this specific exercise, the non-homogeneous part \(5e^{3t}\) suggests using an exponential function as a part of the trial solution.
The trial solution proposed is \(x_p(t) = Ae^{3t}\), where \(A\) is a constant that we need to determine. By comparing the trial solution to the structure of the non-homogeneous term, we ensure the solution aligns with the nature of the differential equation. Here's how the process works:
The trial solution proposed is \(x_p(t) = Ae^{3t}\), where \(A\) is a constant that we need to determine. By comparing the trial solution to the structure of the non-homogeneous term, we ensure the solution aligns with the nature of the differential equation. Here's how the process works:
- Use \(x_p(t) = Ae^{3t}\) as a trial solution.
- Calculate the first derivative \(\frac{\mathrm{d} x_p}{\mathrm{d}t} = 3Ae^{3t}\) and second derivative \(\frac{\mathrm{d}^2 x_p}{\mathrm{d}t^2} = 9Ae^{3t}\).
- Substitute these derivatives into the differential equation.
- Identify and simplify terms to solve for the constant \(A\).