A third-order homogeneous linear equation and three linearly independent solutions are given below. Find a particular solution satisfying the given initial conditions. (3) + 2y" - y'- 2y = 0; y(0) = 1, y'(0) = 4, y''(0) = 0; Y₁ = ex -2x Y₂=eY3=e The particular solution is y(x) = ...
A third-order homogeneous linear equation and three linearly independent solutions are given below. Find a particular solution satisfying the given initial conditions. (3) + 2y" - y'- 2y = 0; y(0) = 1, y'(0) = 4, y''(0) = 0; Y₁ = ex -2x Y₂=eY3=e The particular solution is y(x) = ...
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 26E
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![### Third-Order Homogeneous Linear Equation
#### Problem Statement:
A third-order homogeneous linear equation and three linearly independent solutions are provided below. Find a particular solution that satisfies the given initial conditions.
#### Equation and Initial Conditions:
\[ y^{(3)} + 2y'' - y' - 2y = 0, \]
\[ y(0) = 1, \]
\[ y'(0) = 4, \]
\[ y''(0) = 0. \]
#### Linearly Independent Solutions:
\[ y_1 = e^x, \]
\[ y_2 = e^{-x}, \]
\[ y_3 = e^{-2x}. \]
#### Solution:
To find the particular solution, you need to express the general solution as a linear combination of the independent solutions:
\[ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{-2x}. \]
### Finding Constants (c₁, c₂, c₃):
You will apply the initial conditions to determine the constants \(c_1\), \(c_2\), and \(c_3\). Solving this system will yield the particular solution.
#### The particular solution is:
\[ \boxed{y(x) = \dots} \]
(Note: The exact steps and calculations to find \(c_1\), \(c_2\), and \(c_3\) should be detailed below the problem statement for educational purposes, but based on the image, these steps and calculations are not provided).
This format ensures clarity on an educational website, providing clear steps and separating initial conditions, general solutions, and the particular solution process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbe52e616-7bd7-4d85-a476-7e5fcec53ee5%2F7119e47f-28a4-4347-95aa-23997ef06106%2Fyrplx7g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Third-Order Homogeneous Linear Equation
#### Problem Statement:
A third-order homogeneous linear equation and three linearly independent solutions are provided below. Find a particular solution that satisfies the given initial conditions.
#### Equation and Initial Conditions:
\[ y^{(3)} + 2y'' - y' - 2y = 0, \]
\[ y(0) = 1, \]
\[ y'(0) = 4, \]
\[ y''(0) = 0. \]
#### Linearly Independent Solutions:
\[ y_1 = e^x, \]
\[ y_2 = e^{-x}, \]
\[ y_3 = e^{-2x}. \]
#### Solution:
To find the particular solution, you need to express the general solution as a linear combination of the independent solutions:
\[ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{-2x}. \]
### Finding Constants (c₁, c₂, c₃):
You will apply the initial conditions to determine the constants \(c_1\), \(c_2\), and \(c_3\). Solving this system will yield the particular solution.
#### The particular solution is:
\[ \boxed{y(x) = \dots} \]
(Note: The exact steps and calculations to find \(c_1\), \(c_2\), and \(c_3\) should be detailed below the problem statement for educational purposes, but based on the image, these steps and calculations are not provided).
This format ensures clarity on an educational website, providing clear steps and separating initial conditions, general solutions, and the particular solution process.
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