The Laplace Transform - Differential Equations

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Find the Laplace Transform for the following function.

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

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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$

Find the Laplace Transform for the following function.

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

Tap to see back →

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$

Find the Laplace Transform for the following function.

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

Tap to see back →

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$

Find the Laplace Transform for the following function.

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

Tap to see back →

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$

Find the Laplace transform of the periodic function.

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

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This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

.

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) $\end{Bmatrix}=\frac{1}{1-e^{-sT}$$}\int_$0^T$ $e^{-sT}$f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) $\end{Bmatrix}=\frac{1}{1-e^{-4s}$$}\int_$0^4$ $e^{-st}$S(t)dt$

Find the Laplace transform of the periodic function.

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

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This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

.

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) $\end{Bmatrix}=\frac{1}{1-e^{-sT}$$}\int_$0^T$ $e^{-sT}$f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) $\end{Bmatrix}=\frac{1}{1-e^{-4s}$$}\int_$0^4$ $e^{-st}$S(t)dt$

Find the Laplace transform of the periodic function.

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

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This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

.

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) $\end{Bmatrix}=\frac{1}{1-e^{-sT}$$}\int_$0^T$ $e^{-sT}$f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) $\end{Bmatrix}=\frac{1}{1-e^{-4s}$$}\int_$0^4$ $e^{-st}$S(t)dt$

Find the Laplace transform of the periodic function.

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

Tap to see back →

This particular piecewise function is called a square wave. The period of this function is the length at which it takes the function to return to its starting point.

For this particular function

$S(t)=\left\{\begin{matrix} 1,& 0\leq2\ 0, & 2\leq4 \end{matrix}\right.$

it has a period of

.

and furthermore,

Using the Transform of a Periodic Function Theorem which states,

$\mathcal{L}\begin{Bmatrix} f(t) $\end{Bmatrix}=\frac{1}{1-e^{-sT}$$}\int_$0^T$ $e^{-sT}$f(t)dt$

the problem can be solved as follows.

$\mathcal{L}\begin{Bmatrix} S(t) $\end{Bmatrix}=\frac{1}{1-e^{-4s}$$}\int_$0^4$ $e^{-st}$S(t)dt$