What is the Laplace transform of f(t)= e^(-2t) &nbsp;sin⁡5t?

Question

What is the Laplace transform of f(t)= e^(-2t) &amp;nbsp;sin⁡5t?

star233 · Accepted Answer

To find the Laplace transform of the function

f (t) = e^{- 2 t} \sin (5 t)

, we can use the properties and formulas of Laplace transforms. Let's proceed with the solution.
The Laplace transform of

f (t)

is given by:

ℒ {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t

Substituting the given function

f (t) = e^{- 2 t} \sin (5 t)

, we have:

ℒ {f (t)} = \int_{0}^{\infty} e^{- s t} (e^{- 2 t} \sin (5 t)) d t

To simplify the integral, we can use the property of the Laplace transform:

ℒ {e^{a t} \sin (b t)} = \frac{b}{s^{2} + (b - a)^{2}}

Applying this property, we have:

ℒ {f (t)} = \int_{0}^{\infty} e^{- s t} e^{- 2 t} \sin (5 t) d t

ℒ {f (t)} = \int_{0}^{\infty} e^{- (s + 2) t} \sin (5 t) d t

Comparing this with the property, we can see that

a = - 2

and

b = 5

. Therefore, we can substitute these values into the formula:

ℒ {f (t)} = \frac{5}{(s + 2)^{2} + (5 - (- 2))^{2}}

Simplifying further, we have:

ℒ {f (t)} = \frac{5}{(s + 2)^{2} + 49}

Thus, the Laplace transform of

f (t) = e^{- 2 t} \sin (5 t)

is

\frac{5}{(s + 2)^{2} + 49}

.

What is the Laplace transform of f(t)= e^(-2t)