00 L{t}(s) = | te st dt u = dv = -se^(-st) %3D du = 1dt v = e^(-st) 00 00 -t(e^(-st)/s) e^(-st)/-s^2 (e^(-st))/-s^2 -1/s^2 1/s^2
00 L{t}(s) = | te st dt u = dv = -se^(-st) %3D du = 1dt v = e^(-st) 00 00 -t(e^(-st)/s) e^(-st)/-s^2 (e^(-st))/-s^2 -1/s^2 1/s^2
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
![The image displays the process of finding the Laplace Transform of a function using integration by parts.
1. **Laplace Transform Definition**:
\[
L\{t\}(s) = \int_{0}^{\infty} t e^{-st} dt
\]
2. **Integration by Parts Setup**:
- \( u = t \) and \( dv = -se^{-st} \)
- \( du = 1 \, dt \) and \( v = e^{-st} \)
3. **Integration by Parts Formula**:
\[
\int u \, dv = uv - \int v \, du
\]
4. **Applying the Formula**:
\[
= \left[ -t \frac{e^{-st}}{s} \right]_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{-st}}{s^2} dt
\]
5. **Evaluating Bounds**:
- The first term evaluates to 0 at both limits.
- For the second term:
\[
\int_{0}^{\infty} \frac{e^{-st}}{s^2} dt = \left[ \frac{e^{-st}}{-s^2} \right]_{0}^{\infty}
\]
6. **Final Calculation**:
\[
= 0 - \left(0 - \left( -\frac{1}{s^2} \right)\right) = \frac{1}{s^2}
\]
**Conclusion**: The Laplace Transform of \( t \) is \(\frac{1}{s^2}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf1d0bb8-6fc4-423e-9784-469bf31e55bd%2Fe557b5ab-273e-4b71-a451-3ea548cfa317%2Fjwlxlcj_processed.png&w=3840&q=75)
Transcribed Image Text:The image displays the process of finding the Laplace Transform of a function using integration by parts.
1. **Laplace Transform Definition**:
\[
L\{t\}(s) = \int_{0}^{\infty} t e^{-st} dt
\]
2. **Integration by Parts Setup**:
- \( u = t \) and \( dv = -se^{-st} \)
- \( du = 1 \, dt \) and \( v = e^{-st} \)
3. **Integration by Parts Formula**:
\[
\int u \, dv = uv - \int v \, du
\]
4. **Applying the Formula**:
\[
= \left[ -t \frac{e^{-st}}{s} \right]_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{-st}}{s^2} dt
\]
5. **Evaluating Bounds**:
- The first term evaluates to 0 at both limits.
- For the second term:
\[
\int_{0}^{\infty} \frac{e^{-st}}{s^2} dt = \left[ \frac{e^{-st}}{-s^2} \right]_{0}^{\infty}
\]
6. **Final Calculation**:
\[
= 0 - \left(0 - \left( -\frac{1}{s^2} \right)\right) = \frac{1}{s^2}
\]
**Conclusion**: The Laplace Transform of \( t \) is \(\frac{1}{s^2}\).
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