What is laplacetransform-e^t+1/2 sin(t)+1/2 cos(t) ?

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

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laplacetransform-e^t+1/2 sin(t)+1/2 cos(t)

laplace transform

- e^{t} + \frac{1}{2} sin (t) + \frac{1}{2} cos (t)

\frac{- s ^{2} - 3}{2 ( s - 1 ) ( s ^{2} + 1 )}

Solution steps

L {- e^{t} + \frac{1}{2} sin (t) + \frac{1}{2} cos (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)}

= - L {e^{t}} + \frac{1}{2} L {sin (t)} + \frac{1}{2} L {cos (t)}

L {e^{t}} : \frac{1}{s - 1}

L {sin (t)} : \frac{1}{s ^{2} + 1}

L {cos (t)} : \frac{s}{s ^{2} + 1}

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

Simplify

- \frac{1}{s - 1} + \frac{1}{2} \frac{1}{s ^{2} + 1} + \frac{1}{2} \frac{s}{s ^{2} + 1} : \frac{- s ^{2} - 3}{2 ( s - 1 ) ( s ^{2} + 1 )}

= \frac{- s ^{2} - 3}{2 ( s - 1 ) ( s ^{2} + 1 )}

\int cos (a x + b) d x

\int_{- 1}^{3} 6 x (x - 1) d x

\int csch^{2} (x) d x

x \to - 3 lim (3 x)

\int 5 x + 3 d x

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