22 For exponential decay show that y(t) is the square root of y0) times y(21). How could you find y(31) from y(t) and y(21)?
22 For exponential decay show that y(t) is the square root of y0) times y(21). How could you find y(31) from y(t) and y(21)?
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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![**Question 22:** For exponential decay, show that \( y(1) \) is the square root of \( y(0) \) times \( y(2) \). How could you find \( y(3) \) from \( y(1) \) and \( y(2) \)?
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**Explanation:**
In problems involving exponential decay, the function \( y(t) \) is often expressed in the form \( y(t) = y(0)e^{-kt} \), where \( k \) is the decay constant. You are asked to demonstrate a specific relationship between different values of the function and to determine a method for finding \( y(3) \) using given values of \( y(1) \) and \( y(2) \).
**Step-by-Step Solution:**
1. **Exponential Relationship:**
\[
y(1) = y(0)e^{-k \cdot 1}
\]
\[
y(2) = y(0)e^{-k \cdot 2}
\]
2. **Find \( y(1) \) in terms of \( y(0) \) and \( y(2) \):**
Given:
\[
y(1) = \sqrt{y(0) \cdot y(2)}
\]
Substitute:
\[
\sqrt{y(0) \cdot \left(y(0)e^{-2k}\right)} = \sqrt{y(0)^2 e^{-2k}} = y(0) e^{-k}
\]
So the relationship \( y(1) = \sqrt{y(0) \cdot y(2)} \) holds true based on the definition of exponential decay.
3. **Find \( y(3) \) using \( y(1) \) and \( y(2) \):**
The expression for \( y(t) \) is:
\[
y(t) = y(0)e^{-kt}
\]
Thus:
\[
y(3) = y(0)e^{-3k}
\]
Using \( y(1) \) and \( y(2) \):
\[
y(1) = y(0)e^{-k}, \quad y(2) = y(0)e^{-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F7b3d5440-0e49-42ae-9c05-bf2c46abb9d3%2Fps6x95d_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 22:** For exponential decay, show that \( y(1) \) is the square root of \( y(0) \) times \( y(2) \). How could you find \( y(3) \) from \( y(1) \) and \( y(2) \)?
---
**Explanation:**
In problems involving exponential decay, the function \( y(t) \) is often expressed in the form \( y(t) = y(0)e^{-kt} \), where \( k \) is the decay constant. You are asked to demonstrate a specific relationship between different values of the function and to determine a method for finding \( y(3) \) using given values of \( y(1) \) and \( y(2) \).
**Step-by-Step Solution:**
1. **Exponential Relationship:**
\[
y(1) = y(0)e^{-k \cdot 1}
\]
\[
y(2) = y(0)e^{-k \cdot 2}
\]
2. **Find \( y(1) \) in terms of \( y(0) \) and \( y(2) \):**
Given:
\[
y(1) = \sqrt{y(0) \cdot y(2)}
\]
Substitute:
\[
\sqrt{y(0) \cdot \left(y(0)e^{-2k}\right)} = \sqrt{y(0)^2 e^{-2k}} = y(0) e^{-k}
\]
So the relationship \( y(1) = \sqrt{y(0) \cdot y(2)} \) holds true based on the definition of exponential decay.
3. **Find \( y(3) \) using \( y(1) \) and \( y(2) \):**
The expression for \( y(t) \) is:
\[
y(t) = y(0)e^{-kt}
\]
Thus:
\[
y(3) = y(0)e^{-3k}
\]
Using \( y(1) \) and \( y(2) \):
\[
y(1) = y(0)e^{-k}, \quad y(2) = y(0)e^{-
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