What is laplacetransform e^{3t}sin(4t) ?

The answer to laplacetransform e^{3t}sin(4t) is 4/((s-3)^2+16)

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laplacetransform e^{3t}sin(4t)

laplace transform

e^{3 t} sin (4 t)

\frac{4}{( s - 3 ) ^{2} + 16}

Solution steps

L {e^{3 t} sin (4 t)}

Apply transform rule:

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

then

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

For

e^{3 t} sin (4 t) : f (t) = sin (4 t), a = 3

= L {sin (4 t)} (s - 3)

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

= \frac{4}{( s - 3 ) ^{2} + 16}

\int x^{2} + x d x

n = 0 \sum \infty 9^{n}

f (x) = \frac{x ^{2}}{( x ^{3} ) ( x ^{2} )}

\int \frac{1}{x x ^{2} - 81} d x

f (t) = t^{4} - 20 t^{3} + 112 t^{2}

What is laplacetransform e^{3t}sin(4t) ?
The answer to laplacetransform e^{3t}sin(4t) is 4/((s-3)^2+16)