What is laplacetransform sin(3t)cos(5t) ?

The answer to laplacetransform sin(3t)cos(5t) is 4/(s^2+64)-1/(s^2+4)

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laplacetransform sin(3t)cos(5t)

laplace transform

sin (3 t) cos (5 t)

\frac{4}{s ^{2} + 64} - \frac{1}{s ^{2} + 4}

Solution steps

L {sin (3 t) cos (5 t)}

Rewrite using trig identities

L {\frac{1}{2} sin (8 t) - \frac{1}{2} sin (2 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)}

= \frac{1}{2} L {sin (8 t)} - \frac{1}{2} L {sin (2 t)}

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

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

= \frac{1}{2} \cdot \frac{8}{s ^{2} + 64} - \frac{1}{2} \cdot \frac{2}{s ^{2} + 4}

Refine

\frac{1}{2} \frac{8}{s ^{2} + 64} - \frac{1}{2} \frac{2}{s ^{2} + 4} : \frac{4}{s ^{2} + 64} - \frac{1}{s ^{2} + 4}

= \frac{4}{s ^{2} + 64} - \frac{1}{s ^{2} + 4}

(2 x - 1) d x + (5 y + 8) d y = 0

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

\int_{- 5}^{1} ∣ x + 3 ∣ d x

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} + y^{2})

y^{^{'}} + 9 x^{- 1} y = x^{7}, y (1) = 8

What is laplacetransform sin(3t)cos(5t) ?
The answer to laplacetransform sin(3t)cos(5t) is 4/(s^2+64)-1/(s^2+4)