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 )}

x \to - \frac{π}{2} lim (sec (x))

\frac{d}{d x} ((sin (x) + x)^{2} + 7 cos (5 x))

- \frac{8 x}{x ^{2} + 1}

\frac{d}{d x} (2 x^{2 x} (ln (x) + 1))

\frac{1}{1 + 2 x ^{2}}

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))