What is laplacetransform e^{5t}sin(2t) ?

The answer to laplacetransform e^{5t}sin(2t) is 2/((s-5)^2+4)

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laplacetransform e^{5t}sin(2t)

laplace transform

e^{5 t} sin (2 t)

\frac{2}{( s - 5 ) ^{2} + 4}

Solution steps

L {e^{5 t} sin (2 t)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{5 t} sin (2 t) : f (t) = sin (2 t), a = 5

= L {sin (2 t)} (s - 5)

L {sin (2 t)} : \frac{2}{s ^{2} + 4}

= \frac{2}{( s - 5 ) ^{2} + 4}

\int 25 + x^{2} d x

x \to + 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

y^{^{'}} = - a \cdot y + b \cdot y

f (x) = x^{2} + 2 (3.11)

- 4 cos (6 t)

What is laplacetransform e^{5t}sin(2t) ?
The answer to laplacetransform e^{5t}sin(2t) is 2/((s-5)^2+4)