Question
Asked Jul 8, 2019
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Laplace transfrom.

Evaluate. 

Please see image.

- 2s + 6
s2 4
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- 2s + 6 s2 4

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Expert Answer

Step 1

Given:

 

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J-15-2s+6

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Step 2

Calculation:

Rewrite the given Laplace inverse as follows,

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-2s 6 -2s 6 L s4 s24 2 2 3(2) s222 -2s s22 (2) |s222 2s +3 s2 2s 3sin(2t s2 +22 (s2q-sin(at) 2s = 3sin (2)L (1) s22

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Step 3
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2s Now compute L 22 If LF(s)f(), then L*1 {sF (s)} = f"(f)+f(0) 2 sin (2). So Here L s22 d 2.s (sin (2))(sin (0)) s2 +22 dt =2 cos(2t)0 =2cos(2t) (2)

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