Find the inverse Laplace transform of 7s + 3 s > 0 s2 + 17
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![**Problem:**
Find the inverse Laplace transform of
\[ \frac{7s + 3}{s^2 + 17} \]
for \( s > 0 \).
**Solution:**
To find the inverse Laplace transform, we can separate the expression into two fractions:
\[ \frac{7s}{s^2 + 17} + \frac{3}{s^2 + 17} \]
**Step 1: Inverse Transform of \( \frac{7s}{s^2 + 17} \)**
Recognizing the form \(\frac{s}{s^2 + a^2}\), the inverse Laplace transform is a cosine function:
\[ \mathcal{L}^{-1}\left\{ \frac{s}{s^2 + a^2} \right\} = \cos(at) \]
For our case, \( a^2 = 17 \), hence \( a = \sqrt{17} \). So the inverse Laplace transform of \( \frac{7s}{s^2 + 17} \) is:
\[ 7 \cos(\sqrt{17}t) \]
**Step 2: Inverse Transform of \( \frac{3}{s^2 + 17} \)**
Recognizing the form \(\frac{1}{s^2 + a^2}\), the inverse Laplace transform is a sine function:
\[ \mathcal{L}^{-1}\left\{ \frac{1}{s^2 + a^2} \right\} = \frac{1}{a}\sin(at) \]
Therefore, the inverse Laplace transform of \( \frac{3}{s^2 + 17} \) is:
\[ \frac{3}{\sqrt{17}} \sin(\sqrt{17} t) \]
**Final Solution:**
Combining both parts, the inverse Laplace transform of the original function is:
\[ 7 \cos(\sqrt{17}t) + \frac{3}{\sqrt{17}} \sin(\sqrt{17}t) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb738ee4b-0d92-438d-b96f-d8d02fb325c0%2Fbc65d296-c0e6-4eca-80e0-0cb2bdd396bf%2F540ezb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find the inverse Laplace transform of
\[ \frac{7s + 3}{s^2 + 17} \]
for \( s > 0 \).
**Solution:**
To find the inverse Laplace transform, we can separate the expression into two fractions:
\[ \frac{7s}{s^2 + 17} + \frac{3}{s^2 + 17} \]
**Step 1: Inverse Transform of \( \frac{7s}{s^2 + 17} \)**
Recognizing the form \(\frac{s}{s^2 + a^2}\), the inverse Laplace transform is a cosine function:
\[ \mathcal{L}^{-1}\left\{ \frac{s}{s^2 + a^2} \right\} = \cos(at) \]
For our case, \( a^2 = 17 \), hence \( a = \sqrt{17} \). So the inverse Laplace transform of \( \frac{7s}{s^2 + 17} \) is:
\[ 7 \cos(\sqrt{17}t) \]
**Step 2: Inverse Transform of \( \frac{3}{s^2 + 17} \)**
Recognizing the form \(\frac{1}{s^2 + a^2}\), the inverse Laplace transform is a sine function:
\[ \mathcal{L}^{-1}\left\{ \frac{1}{s^2 + a^2} \right\} = \frac{1}{a}\sin(at) \]
Therefore, the inverse Laplace transform of \( \frac{3}{s^2 + 17} \) is:
\[ \frac{3}{\sqrt{17}} \sin(\sqrt{17} t) \]
**Final Solution:**
Combining both parts, the inverse Laplace transform of the original function is:
\[ 7 \cos(\sqrt{17}t) + \frac{3}{\sqrt{17}} \sin(\sqrt{17}t) \]
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