Question

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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