What is laplacetransform f(t)=5cos(8t)+3e^{-4}sin(3t) ?

The answer to laplacetransform f(t)=5cos(8t)+3e^{-4}sin(3t) is (5s)/(s^2+64)+9/(e^4(s^2+9))

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laplacetransform f(t)=5cos(8t)+3e^{-4}sin(3t)

laplace transform

f (t) = 5 cos (8 t) + 3 e^{- 4} sin (3 t)

\frac{5 s}{s ^{2} + 64} + \frac{9}{e ^{4} ( s ^{2} + 9 )}

Solution steps

L {5 cos (8 t) + 3 e^{- 4} sin (3 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= 5 L {cos (8 t)} + 3 e^{- 4} L {sin (3 t)}

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

= 5 \cdot \frac{s}{s ^{2} + 64} + 3 e^{- 4} \frac{3}{s ^{2} + 9}

Refine

5 \frac{s}{s ^{2} + 64} + 3 e^{- 4} \frac{3}{s ^{2} + 9} : \frac{5 s}{s ^{2} + 64} + \frac{9}{e ^{4} ( s ^{2} + 9 )}

= \frac{5 s}{s ^{2} + 64} + \frac{9}{e ^{4} ( s ^{2} + 9 )}

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What is laplacetransform f(t)=5cos(8t)+3e^{-4}sin(3t) ?
The answer to laplacetransform f(t)=5cos(8t)+3e^{-4}sin(3t) is (5s)/(s^2+64)+9/(e^4(s^2+9))