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

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

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

laplace transform

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

2 e^{2 t}

y = x - 2, x = y^{2}

\int_{- 1}^{1} (1 - x^{2})^{\frac{3}{2}} d x

\frac{x}{6 x - 4}, at x = 9

\frac{\partial}{\partial x} (2 x^{4} y z^{3})

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