Step 1 We have to form Laplace transform integral first. Step 2 We have f(t)=etcos⁡t We know that, F(s)=∫0∞e−stf(t)dt ⇒F(s)=∫0∞cos⁡(t)e(1−s)tdt by integrath by pcuts ⇒F(s)=[cos⁡(t)e(1−s)t1−s]0∞+∫(e(1−s)t1−s)sin⁡tdt ⇒F(s)=−11−s+11−s[e(1−s)tsin⁡t1−s]0∞−∫(e(1−s)t1−s)cos⁡tdt ⇒F(s)=−11−s−1(1−s)2 ⇒F(s)[1+1(1−s)2]=−1(1−s) ⇒F(s)=−(1−s)2(1−s)[(1−s)2+1] ⇒F(s)=s−1[(s−1)2+1] Answer: F(s)=s−1[(s−1)2+1]

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Name: Solve f(t)=etcos⁡t
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve f(t)=etcos⁡t

UkusakazaL

Answered question

2020-11-01

Solve

f (t) = e^{t} \cos t

Answer & Explanation

averes8

Skilled2020-11-02Added 92 answers

Step 1
We have to form Laplace transform integral first.
Step 2
We have

f (t) = e^{t} \cos t

We know that,

F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t

\Rightarrow F (s) = \int_{0}^{\infty} \cos (t) e^{(1 - s) t} d t

by integrath by pcuts

\Rightarrow F (s) = {[\frac{\cos (t) e^{(1 - s) t}}{1 - s}]}_{0}^{\infty} + \int (\frac{e^{(1 - s) t}}{1 - s}) \sin t d t

\Rightarrow F (s) = - \frac{1}{1 - s} + \frac{1}{1 - s} {[\frac{e^{(1 - s) t} \sin t}{1 - s}]}_{0}^{\infty} - \int (\frac{e^{(1 - s) t}}{1 - s}) \cos t d t

\Rightarrow F (s) = - \frac{1}{1 - s} - \frac{1}{(1 - s)^{2}}

\Rightarrow F (s) [1 + \frac{1}{(1 - s)^{2}}] = - \frac{1}{(1 - s)}

\Rightarrow F (s) = \frac{- (1 - s)^{2}}{(1 - s) [(1 - s)^{2} + 1]}

\Rightarrow F (s) = \frac{s - 1}{[(s - 1)^{2} + 1]}

Answer:

F (s) = \frac{s - 1}{[(s - 1)^{2} + 1]}

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