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Use the method of Variation of Parameters to solve. SHOW WORK. у" - у' — 2у %3D 2е-t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving Differential Equation using Variation of Parameters
**Problem Statement:**
Use the method of Variation of Parameters to solve. SHOW WORK.
\[ y'' - y' - 2y = 2e^{-t} \]
**Solution:**
The given second-order linear differential equation is:
\[ y'' - y' - 2y = 2e^{-t} \]
#### Step 1: Solve the Homogeneous Equation
First, solve the homogeneous equation associated with the given differential equation:
\[ y'' - y' - 2y = 0 \]
The characteristic equation is:
\[ r^2 - r - 2 = 0 \]
Factoring gives:
\[ (r - 2)(r + 1) = 0 \]
So, the roots are \( r_1 = 2 \) and \( r_2 = -1 \).
The general solution to the homogeneous equation is:
\[ y_h(t) = C_1 e^{2t} + C_2 e^{-t} \]
#### Step 2: Determine Particular Solution using Variation of Parameters
To find a particular solution \( y_p(t) \), assume:
\[ y_p(t) = u_1(t) e^{2t} + u_2(t) e^{-t} \]
where \( u_1(t) \) and \( u_2(t) \) are functions to be determined.
The formulas for \( u_1(t) \) and \( u_2(t) \) in Variation of Parameters are:
\[ u_1'(t) = -\frac{y_2(t) g(t)}{W(y_1, y_2)} \]
\[ u_2'(t) = \frac{y_1(t) g(t)}{W(y_1, y_2)} \]
where:
- \( g(t) = 2e^{-t} \)
- \( y_1(t) = e^{2t} \)
- \( y_2(t) = e^{-t} \)
- \( W(y_1, y_2) \) is the Wronskian of \( y_1 \) and \( y_2 \):
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff139fb70-ebd9-481c-a8b8-1b8ffb3b31f4%2Fafffee5d-9894-4dd9-8aa4-d76366d52080%2Fejeiutq_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Differential Equation using Variation of Parameters
**Problem Statement:**
Use the method of Variation of Parameters to solve. SHOW WORK.
\[ y'' - y' - 2y = 2e^{-t} \]
**Solution:**
The given second-order linear differential equation is:
\[ y'' - y' - 2y = 2e^{-t} \]
#### Step 1: Solve the Homogeneous Equation
First, solve the homogeneous equation associated with the given differential equation:
\[ y'' - y' - 2y = 0 \]
The characteristic equation is:
\[ r^2 - r - 2 = 0 \]
Factoring gives:
\[ (r - 2)(r + 1) = 0 \]
So, the roots are \( r_1 = 2 \) and \( r_2 = -1 \).
The general solution to the homogeneous equation is:
\[ y_h(t) = C_1 e^{2t} + C_2 e^{-t} \]
#### Step 2: Determine Particular Solution using Variation of Parameters
To find a particular solution \( y_p(t) \), assume:
\[ y_p(t) = u_1(t) e^{2t} + u_2(t) e^{-t} \]
where \( u_1(t) \) and \( u_2(t) \) are functions to be determined.
The formulas for \( u_1(t) \) and \( u_2(t) \) in Variation of Parameters are:
\[ u_1'(t) = -\frac{y_2(t) g(t)}{W(y_1, y_2)} \]
\[ u_2'(t) = \frac{y_1(t) g(t)}{W(y_1, y_2)} \]
where:
- \( g(t) = 2e^{-t} \)
- \( y_1(t) = e^{2t} \)
- \( y_2(t) = e^{-t} \)
- \( W(y_1, y_2) \) is the Wronskian of \( y_1 \) and \( y_2 \):
\[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix
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