Lleft{t-e^{-3t}right}] which of the laplace transform is 1.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}+frac{1}{s-3} 2.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}-frac{1}{s-3} 3.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}+frac{1}{s+3} 4.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}-frac{1}{s+3}

Question
Laplace transform
asked 2020-12-25
\(L\left\{t-e^{-3t}\right\}\]
which of the laplace transform is
\(1.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s-3}\)
\(2.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s-3}\)
\(3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}\)
\(4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}\)

Answers (1)

2020-12-26
Step 1
We know the Laplace transforms \(L\left\{t^{n}\right\}=\frac{n!}{s^{n+1}},s>0\)
\(L\left\{e^{at}\right\}=\frac{1}{(s-a)},s>0\)
\(\text{Now, }L\left\{t-e^{-3t}\right\}=L\left\{t\right\}-L\left\{e^{-3t}\right\}\)
\(L\left\{t\right\}=\frac{1}{s^{2}},s>0\)
\(L\left\{t-e^{-3t}\right\}=-\frac{1}{s+3},s>-3\)
\(L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{(s+3)}\)
Step 2
Answer: \(L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}\) option 4) is correct
0

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