Question

Find the Laplace transform of
Foto
e-2 (3 cos 6t - 5 sin 6t)
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Answer 1

Answer:

thanks for free points

Answer 2

"It seems that this what you are looking for"

Find the Laplace transform of f(t) = e^{-2t}(3cos6t-5sin6t).

Answer:

The Laplace transform of given function f(t) is equal to  $\frac{3s-24}{(s+2)^{2}+\beta ^{2}}$.

Explanation:

We have given the expression is :

$f(t)= e^{-2t}(3cos6t-5sin6t)$

The symbol 'L' is Laplace Transform operator. The given function f depends upon 't' and its Laplace transform L(f(t)) will depend upon 's'. The parameter 's' is taken as real positive number.

Write the Laplace transform of some standard functions using First shifting theorem:

$L(e^{\alpha t} {sin\beta t)=\frac{\beta }{(s-\alpha )^{2}+\beta ^{2}}$                                         ...............(1)

$L(e^{\alpha t} {cos\beta t)=\frac{ s-\alpha }{(s-\alpha )^{2}+\beta ^{2}}$                                         ................(2)

Now simplify the given expression:

$f(t)= 3(e^{-2t}cos6t)-5(e^{-2t}sin6t)$

Apply the Laplace transformation:

$L(f(t))= L[3(e^{-2t}cos6t)-5(e^{-2t}sin6t)]$

$L(f(t))= 3L((e^{-2t}cos6t))-5L((e^{-2t}sin6t))$

By using the equation (1) and (2) and put the value of α = -2 and β = 6 ,$L(f(t))=3\big[\frac{s-(-2)}{(s-(-2))^{2}+\beta ^{2}} \big] -5\big[\frac{6}{(s-(-2))^{2}+\beta ^{2}} \big]$

$L(f(t))=3\big[\frac{(s+2)}{(s+2)^{2}+\beta ^{2}} \big] -5\big[\frac{6}{(s+2)^{2}+\beta ^{2}} \big]$

$L(f(t))=\big[\frac{3s+6}{(s+2)^{2}+\beta ^{2}} \big] -\big[\frac{30}{(s+2)^{2}+\beta ^{2}} \big]$

$L(f(t))=\frac{3s+6-30}{(s+2)^{2}+\beta ^{2}}$

$L(f(t))=\frac{3s-24}{(s+2)^{2}+\beta ^{2}}$

Hence the result of given function f(t) is obtained.