correct answer. I have attached the correct answer in this question. question: solve the given initial-value problem.sketch the graph of both the forcing function and the solution. y″ + 2y′ +5y = 10 –10u(t – π) y(0)= 0 y′(0) = 0
correct answer. I have attached the correct answer in this question. question: solve the given initial-value problem.sketch the graph of both the forcing function and the solution. y″ + 2y′ +5y = 10 –10u(t – π) y(0)= 0 y′(0) = 0
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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please help show step on how to get the correct answer. I have attached the correct answer in this question.
question:
solve the given initial-value problem.sketch the graph
of both the forcing function and the solution.
y″ + 2y′ +5y = 10 –10u(t – π)
y(0)= 0
y′(0) = 0
![### Mathematical Expression
The expression shown is:
\[
\frac{5}{2} \Big( 1 - e^{-t}(\cos 2t + \sin 2t) \Big) - u(t - \pi) \Big[ 1 - e^{-(t - \pi)} (\cos 2t + \sin 2t) \Big]
\]
Where:
- \( e \) represents the base of the natural logarithm.
- \( t \) is a variable, typically representing time.
- \( \cos \) and \( \sin \) are trigonometric functions representing cosine and sine, respectively.
- \( u(t - \pi) \) is the unit step function, which is 0 for \( t < \pi \) and 1 for \( t \geq \pi \).
### Detailed Explanation
The expression can be broken down into two main components:
1. **First Component:**
\[
\frac{5}{2} \Big( 1 - e^{-t}(\cos 2t + \sin 2t) \Big)
\]
- Represents a modulated response that starts from an initial value and asymptotically approaches a steady state as \( t \) increases.
- The exponential decay \( e^{-t} \) modifies the amplitude of the trigonometric functions, influencing the oscillatory behavior of the system.
2. **Second Component:**
\[
- u(t - \pi) \Big[ 1 - e^{-(t - \pi)} (\cos 2t + \sin 2t) \Big]
\]
- Introduces a shift in the system's behavior at \( t = \pi \).
- The unit step function \( u(t - \pi) \) activates this part of the expression only when \( t \) reaches \( \pi \) or later.
- Similar to the first component, but begins its effect at \( t = \pi \).
This expression might be used in contexts such as signals and systems, control theory, or differential equations to model transient responses in physical systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe31b6b9e-ba6c-4dca-9489-09cb6373ae58%2F4f85b852-5f56-4239-9a76-1dead0e87fa1%2Fuunw7xq_processed.png&w=3840&q=75)
Transcribed Image Text:### Mathematical Expression
The expression shown is:
\[
\frac{5}{2} \Big( 1 - e^{-t}(\cos 2t + \sin 2t) \Big) - u(t - \pi) \Big[ 1 - e^{-(t - \pi)} (\cos 2t + \sin 2t) \Big]
\]
Where:
- \( e \) represents the base of the natural logarithm.
- \( t \) is a variable, typically representing time.
- \( \cos \) and \( \sin \) are trigonometric functions representing cosine and sine, respectively.
- \( u(t - \pi) \) is the unit step function, which is 0 for \( t < \pi \) and 1 for \( t \geq \pi \).
### Detailed Explanation
The expression can be broken down into two main components:
1. **First Component:**
\[
\frac{5}{2} \Big( 1 - e^{-t}(\cos 2t + \sin 2t) \Big)
\]
- Represents a modulated response that starts from an initial value and asymptotically approaches a steady state as \( t \) increases.
- The exponential decay \( e^{-t} \) modifies the amplitude of the trigonometric functions, influencing the oscillatory behavior of the system.
2. **Second Component:**
\[
- u(t - \pi) \Big[ 1 - e^{-(t - \pi)} (\cos 2t + \sin 2t) \Big]
\]
- Introduces a shift in the system's behavior at \( t = \pi \).
- The unit step function \( u(t - \pi) \) activates this part of the expression only when \( t \) reaches \( \pi \) or later.
- Similar to the first component, but begins its effect at \( t = \pi \).
This expression might be used in contexts such as signals and systems, control theory, or differential equations to model transient responses in physical systems.
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