What is laplacetransform f(t)=6e^{-5t}+e^{3t}+5t^3-9 ?

The answer to laplacetransform f(t)=6e^{-5t}+e^{3t}+5t^3-9 is 6/(s+5)+1/(s-3)+(30)/(s^4)-9/s

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

laplacetransform f(t)=6e^{-5t}+e^{3t}+5t^3-9

laplace transform

f (t) = 6 e^{- 5 t} + e^{3 t} + 5 t^{3} - 9

\frac{6}{s + 5} + \frac{1}{s - 3} + \frac{30}{s ^{4}} - \frac{9}{s}

Solution steps

L {6 e^{- 5 t} + e^{3 t} + 5 t^{3} - 9}

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)}

= 6 L {e^{- 5 t}} + L {e^{3 t}} + 5 L {t^{3}} - L {9}

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

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

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

L {9} : \frac{9}{s}

= 6 \cdot \frac{1}{s + 5} + \frac{1}{s - 3} + 5 \cdot \frac{6}{s ^{4}} - \frac{9}{s}

Refine

6 \frac{1}{s + 5} + \frac{1}{s - 3} + 5 \frac{6}{s ^{4}} - \frac{9}{s} : \frac{6}{s + 5} + \frac{1}{s - 3} + \frac{30}{s ^{4}} - \frac{9}{s}

= \frac{6}{s + 5} + \frac{1}{s - 3} + \frac{30}{s ^{4}} - \frac{9}{s}

What is laplacetransform f(t)=6e^{-5t}+e^{3t}+5t^3-9 ?
The answer to laplacetransform f(t)=6e^{-5t}+e^{3t}+5t^3-9 is 6/(s+5)+1/(s-3)+(30)/(s^4)-9/s