Find the general solution for y" + 4y = sin(2t)
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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![## General Solution of the Differential Equation
**Problem Statement:**
Find the general solution for
\[ y'' + 4y = \sin(2t) \]
**Solution:**
To solve this second-order non-homogeneous linear differential equation, we need to find the general solution, which is a combination of the homogeneous solution and a particular solution.
### 1. Homogeneous Solution
First, we solve the corresponding homogeneous equation:
\[ y'' + 4y = 0 \]
The characteristic equation is:
\[ r^2 + 4 = 0 \]
Solving for \(r\):
\[ r^2 = -4 \]
\[ r = \pm 2i \]
Thus, the general solution to the homogeneous equation is:
\[ y_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \]
where \(C_1\) and \(C_2\) are arbitrary constants.
### 2. Particular Solution
Next, we find a particular solution to the non-homogeneous equation. Considering the form of the non-homogeneous term (\(\sin(2t)\)), we can assume a particular solution of the form:
\[ y_p(t) = A t \cos(2t) + B t \sin(2t) \]
Substituting \(y_p(t)\), its first, and second derivatives into the original differential equation and solving for constants \(A\) and \(B\), we obtain the particular solution.
### 3. General Solution
The general solution is then the sum of the homogeneous solution and the particular solution:
\[ y(t) = y_h(t) + y_p(t) \]
\[ y(t) = C_1 \cos(2t) + C_2 \sin(2t) + y_p(t) \]
### Conclusion
Therefore, the general solution of the given differential equation is:
\[ y(t) = C_1 \cos(2t) + C_2 \sin(2t) + y_p(t) \]
where \(C_1\) and \(C_2\) are constants determined by initial conditions, and \(y_p(t)\) is any particular solution found by substitution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F946e5b6f-098d-47d1-a6b1-15872ebd1054%2F05684e0e-005d-4688-ab48-394c665dbc1d%2F335fct_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## General Solution of the Differential Equation
**Problem Statement:**
Find the general solution for
\[ y'' + 4y = \sin(2t) \]
**Solution:**
To solve this second-order non-homogeneous linear differential equation, we need to find the general solution, which is a combination of the homogeneous solution and a particular solution.
### 1. Homogeneous Solution
First, we solve the corresponding homogeneous equation:
\[ y'' + 4y = 0 \]
The characteristic equation is:
\[ r^2 + 4 = 0 \]
Solving for \(r\):
\[ r^2 = -4 \]
\[ r = \pm 2i \]
Thus, the general solution to the homogeneous equation is:
\[ y_h(t) = C_1 \cos(2t) + C_2 \sin(2t) \]
where \(C_1\) and \(C_2\) are arbitrary constants.
### 2. Particular Solution
Next, we find a particular solution to the non-homogeneous equation. Considering the form of the non-homogeneous term (\(\sin(2t)\)), we can assume a particular solution of the form:
\[ y_p(t) = A t \cos(2t) + B t \sin(2t) \]
Substituting \(y_p(t)\), its first, and second derivatives into the original differential equation and solving for constants \(A\) and \(B\), we obtain the particular solution.
### 3. General Solution
The general solution is then the sum of the homogeneous solution and the particular solution:
\[ y(t) = y_h(t) + y_p(t) \]
\[ y(t) = C_1 \cos(2t) + C_2 \sin(2t) + y_p(t) \]
### Conclusion
Therefore, the general solution of the given differential equation is:
\[ y(t) = C_1 \cos(2t) + C_2 \sin(2t) + y_p(t) \]
where \(C_1\) and \(C_2\) are constants determined by initial conditions, and \(y_p(t)\) is any particular solution found by substitution.
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