In the previous step we determined the piecewise continuous function f(t). So, f(t) = |2t – 2, 0 1 We now use this to find the Laplace transform L{f(t)} e-stret) dt. Since f is defined in two pieces, L{(t)} is expressed as the sum of two integrals. L{f(t)} -strce) dt 00 dt + 2t - dt = 0 + (t – 1)e-st dt
In the previous step we determined the piecewise continuous function f(t). So, f(t) = |2t – 2, 0 1 We now use this to find the Laplace transform L{f(t)} e-stret) dt. Since f is defined in two pieces, L{(t)} is expressed as the sum of two integrals. L{f(t)} -strce) dt 00 dt + 2t - dt = 0 + (t – 1)e-st dt
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
Please write legibly to avoid confusion of final answer
![In the previous step we determined the piecewise continuous function f(t).
f(t) =
2t
So,
0 <ts 1
t > 1
2,
We now use this to find the Laplace transform L{f{t)} =
-strt) dt. Since f is defined in two pieces,
e
L{f(t)} is expressed as the sum of two integrals.
L{r(t)}
e-st(t) dt
dt +
dt
= 0 +
(t - 1)e-st dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5085ac77-62eb-41f7-a43e-d213f0473bb9%2Fc4798e7c-9b4e-4252-87bc-7c56b1ec1e9e%2Fy1sv3s_processed.png&w=3840&q=75)
Transcribed Image Text:In the previous step we determined the piecewise continuous function f(t).
f(t) =
2t
So,
0 <ts 1
t > 1
2,
We now use this to find the Laplace transform L{f{t)} =
-strt) dt. Since f is defined in two pieces,
e
L{f(t)} is expressed as the sum of two integrals.
L{r(t)}
e-st(t) dt
dt +
dt
= 0 +
(t - 1)e-st dt
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