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 \frac{1}{y ( y ^{2} + 1 )} d y

\int \frac{25 - 4 x ^{2}}{x} d x

\int_{2}^{\infty} \frac{1}{x ^{2} + 4} d x

\frac{y ^{3}}{18} + \frac{3}{2 y}

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