What is laplacetransform t-2+e^{-5t}-sin(5t)+1.5 ?

The answer to laplacetransform t-2+e^{-5t}-sin(5t)+1.5 is 1/(s^2)-(0.5)/s+1/(s+5)-5/(s^2+25)

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laplacetransform t-2+e^{-5t}-sin(5t)+1.5

laplace transform

t - 2 + e^{- 5 t} - sin (5 t) + 1.5

\frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

Solution steps

L {t - 2 + e^{- 5 t} - sin (5 t) + 1.5}

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 {t} - L {2} + L {e^{- 5 t}} - L {sin (5 t)} + L {1.5}

L {t} : \frac{1}{s ^{2}}

L {2} : \frac{2}{s}

L {e^{- 5 t}} : \frac{1}{s + 5}

L {sin (5 t)} : \frac{5}{s ^{2} + 25}

L {1.5} : \frac{1.5}{s}

= \frac{1}{s ^{2}} - \frac{2}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25} + \frac{1.5}{s}

Refine

\frac{1}{s ^{2}} - \frac{2}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25} + \frac{1.5}{s} : \frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

= \frac{1}{s ^{2}} - \frac{0.5}{s} + \frac{1}{s + 5} - \frac{5}{s ^{2} + 25}

x sin (2 x), 0

f (x) = 5 - 2 x^{2}, (- 2, - 3)

x \to \infty lim (- 2 x + \frac{1}{x})

- 3 cos (5 t)

\frac{x - 5}{x - 2}

What is laplacetransform t-2+e^{-5t}-sin(5t)+1.5 ?
The answer to laplacetransform t-2+e^{-5t}-sin(5t)+1.5 is 1/(s^2)-(0.5)/s+1/(s+5)-5/(s^2+25)