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

The answer to laplacetransform e^{-t}+2e^{-2t}+te^{-3t} is 1/(s+1)+2/(s+2)+1/((s+3)^2)

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laplacetransform e^{-t}+2e^{-2t}+te^{-3t}

laplace transform

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

\frac{1}{s + 1} + \frac{2}{s + 2} + \frac{1}{( s + 3 ) ^{2}}

Solution steps

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

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

L {e^{- 2 t}} : \frac{1}{s + 2}

L {t e^{- 3 t}} : \frac{1}{( s + 3 ) ^{2}}

= \frac{1}{s + 1} + 2 \cdot \frac{1}{s + 2} + \frac{1}{( s + 3 ) ^{2}}

Refine

\frac{1}{s + 1} + 2 \frac{1}{s + 2} + \frac{1}{( s + 3 ) ^{2}} : \frac{1}{s + 1} + \frac{2}{s + 2} + \frac{1}{( s + 3 ) ^{2}}

= \frac{1}{s + 1} + \frac{2}{s + 2} + \frac{1}{( s + 3 ) ^{2}}

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