Definition of Laplace Transform - Differential Equations

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Question

Find the Laplace Transform for the following function.

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

Answer

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,

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

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

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

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

Compare your answer with the correct one above

Question

Find the Laplace Transform for the following function.

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

Answer

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,

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

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

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

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

Compare your answer with the correct one above

Question

Find the Laplace Transform for the following function.

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

Answer

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,

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

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

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

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

Compare your answer with the correct one above

Question

Find the Laplace Transform for the following function.

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

Answer

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,

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

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

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

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

Compare your answer with the correct one above

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