9) Here fct) - 2lt+ 3) e 4) (++5) NOw, L{t+s} = L{ t}t 3 L {13 %3D 3 = g(s) say L{21++3}e-(t+5)} So by first tromlation theorem] - 2e-5 - 2e-5 r: gls) = + There fore the Laplaee trans form of f4) = 2(4+3) e (++5) s 2eslrt %3D 3
9) Here fct) - 2lt+ 3) e 4) (++5) NOw, L{t+s} = L{ t}t 3 L {13 %3D 3 = g(s) say L{21++3}e-(t+5)} So by first tromlation theorem] - 2e-5 - 2e-5 r: gls) = + There fore the Laplaee trans form of f4) = 2(4+3) e (++5) s 2eslrt %3D 3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
(4) Can you translate this into words? Please explain the solution in detail because I'm going to make a script out of it.
![4)
Here f(t) - 2ltt 3) e
(t+5)
No cw, L{t+3} = 2{ t}t 3L{13
+ 3
g(s) say
%3D
3ト 3
-し2
- 2e-5
g(s+1) by first romlation theoremi]
gls+1)
- 2e-5)
(3+リ
There fore the laplaee trans form of
J4) = a (4+3) e"(t+9) s z0° [
ト3
ろ+」](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdde3ae2a-42ac-433e-9f60-e7d93b0f2f12%2Fd4ff11a6-0147-48d2-85d2-18ae08842688%2Fczcmqk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4)
Here f(t) - 2ltt 3) e
(t+5)
No cw, L{t+3} = 2{ t}t 3L{13
+ 3
g(s) say
%3D
3ト 3
-し2
- 2e-5
g(s+1) by first romlation theoremi]
gls+1)
- 2e-5)
(3+リ
There fore the laplaee trans form of
J4) = a (4+3) e"(t+9) s z0° [
ト3
ろ+」
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