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Differential Equations : Definition of Laplace Transform

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Example Question #82 : Differential Equations

Find the Laplace Transform for the following function.

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}$

Possible Answers:

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s<-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+3}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s-5}, s>-5$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

Correct answer:

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\frac{1}{s+5}, s>-5$

Explanation:

To find the Laplace Transform for the following function

$\mathcal{L}\begin{Bmatrix} e^{-5t}\end{Bmatrix}$

use the transforms for basic functions which state,

$\mathcal{L}\begin{Bmatrix} 1 \end{Bmatrix}=\frac{1}{s}$

$\mathcal{L}\begin{Bmatrix} e^{at} \end{Bmatrix}=\frac{1}{s-a}$

$\mathcal{L}\begin{Bmatrix} e^{-5t} \end{Bmatrix}=\int_0^\infty e^{-st}e^{-5t}dt=\int_0^\infty e^{-(s+5)t} dt$

$\\=\frac{-e^{-(s+5)t}}{s+5}\bigg|_0^\infty \\\\=\frac{1}{s+5}, s>-5$

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Differential Equations : Definition of Laplace Transform

Study concepts, example questions & explanations for Differential Equations

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Example Question #82 : Differential Equations

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