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# f(t)=3e^{2t} Determine L[f] Let f be a function defined on an interval [0,infty) The Laplace transform of f is the function F(s) defined by F(s) =int_0^infty e^{-st}f(t)dt provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].

Laplace transform
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asked 2021-01-08
$$f(t)=3e^{2t}$$
Determine L[f]
Let f be a function defined on an interval $$[0,\infty)$$
The Laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].

## Answers (1)

2021-01-09
Step 1
we have to find the laplace of $$f(t)=3e^{2t}$$ by using the definition of the laplace transform.
as we know that the laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
therefore,
$$L[f]=\int_0^\infty e^{-st}f(t)dt$$
$$L[3e^{2t}]=\int_0^\infty e^{-st}(3e^{2t})dt$$
$$=3\int_0^\infty e^{-st}(e^{2t})dt$$
$$=3\int_0^\infty e^{-st+2t}dt$$
$$=3\int_0^\infty e^{-(s-2)t}dt$$
Step 2
now as we know that $$\int e^{bx}dx=\frac{e^{bx}}{b}$$
therefore,
$$L[3e^{2t}]=3\left(\frac{e^{-(s-2)t}}{-(s-2)}\right)_0^\infty$$
$$=-\frac{3}{(s-2)}\left(e^{-(s-2)t}\right)_0^\infty$$
$$=-\frac{3}{(s-2)}\left(e^{-(s-2)\infty}-e^{-(s-2)(0)}\right)$$
$$=-\frac{3}{(s-2)}(e^{\infty}-e^0)$$
$$=-\frac{3}{(s-2)}(0-1)$$
$$=\frac{3}{(s-2)}$$
Step 3
therefore the laplace transform of $$f(t)=3e^{2t} \text{ is } \frac{3}{(s-2)}$$

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