What is laplacetransform e^{-3t}+3+2t+4t^2+5t^3 ?

The answer to laplacetransform e^{-3t}+3+2t+4t^2+5t^3 is 1/(s+3)+3/s+2/(s^2)+8/(s^3)+(30)/(s^4)

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laplacetransform e^{-3t}+3+2t+4t^2+5t^3

laplace transform

e^{- 3 t} + 3 + 2 t + 4 t^{2} + 5 t^{3}

\frac{1}{s + 3} + \frac{3}{s} + \frac{2}{s ^{2}} + \frac{8}{s ^{3}} + \frac{30}{s ^{4}}

Solution steps

L {e^{- 3 t} + 3 + 2 t + 4 t^{2} + 5 t^{3}}

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^{- 3 t}} + L {3} + 2 L {t} + 4 L {t^{2}} + 5 L {t^{3}}

L {e^{- 3 t}} : \frac{1}{s + 3}

L {3} : \frac{3}{s}

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

L {t^{2}} : \frac{2}{s ^{3}}

L {t^{3}} : \frac{6}{s ^{4}}

= \frac{1}{s + 3} + \frac{3}{s} + 2 \cdot \frac{1}{s ^{2}} + 4 \cdot \frac{2}{s ^{3}} + 5 \cdot \frac{6}{s ^{4}}

Refine

\frac{1}{s + 3} + \frac{3}{s} + 2 \frac{1}{s ^{2}} + 4 \frac{2}{s ^{3}} + 5 \frac{6}{s ^{4}} : \frac{1}{s + 3} + \frac{3}{s} + \frac{2}{s ^{2}} + \frac{8}{s ^{3}} + \frac{30}{s ^{4}}

= \frac{1}{s + 3} + \frac{3}{s} + \frac{2}{s ^{2}} + \frac{8}{s ^{3}} + \frac{30}{s ^{4}}

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

\int cos (\frac{x}{2}) d x

f (x) = 5 x - x^{2}, (1, 4)

(1 + x) d y - y d x = 0

x \to \infty lim (6^{x})

What is laplacetransform e^{-3t}+3+2t+4t^2+5t^3 ?
The answer to laplacetransform e^{-3t}+3+2t+4t^2+5t^3 is 1/(s+3)+3/s+2/(s^2)+8/(s^3)+(30)/(s^4)