What is laplacetransform (1+te^{-t})^3 ?

The answer to laplacetransform (1+te^{-t})^3 is 1/s+3/((s+1)^2)+6/((s+2)^3)+6/((s+3)^4)

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laplacetransform (1+te^{-t})^3

laplace transform

(1 + t e^{- t})^{3}

\frac{1}{s} + \frac{3}{( s + 1 ) ^{2}} + \frac{6}{( s + 2 ) ^{3}} + \frac{6}{( s + 3 ) ^{4}}

Solution steps

L {(1 + t e^{- t})^{3}}

Expand

(1 + t e^{- t})^{3} : 1 + 3 e^{- t} t + 3 e^{- 2 t} t^{2} + e^{- 3 t} t^{3}

= L {1 + 3 e^{- t} t + 3 e^{- 2 t} t^{2} + e^{- 3 t} 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 {1} + 3 L {e^{- t} t} + 3 L {e^{- 2 t} t^{2}} + L {e^{- 3 t} t^{3}}

L {1} : \frac{1}{s}

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

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

L {e^{- 3 t} t^{3}} : \frac{6}{( s + 3 ) ^{4}}

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

Refine

\frac{1}{s} + 3 \frac{1}{( s + 1 ) ^{2}} + 3 \frac{2}{( s + 2 ) ^{3}} + \frac{6}{( s + 3 ) ^{4}} : \frac{1}{s} + \frac{3}{( s + 1 ) ^{2}} + \frac{6}{( s + 2 ) ^{3}} + \frac{6}{( s + 3 ) ^{4}}

= \frac{1}{s} + \frac{3}{( s + 1 ) ^{2}} + \frac{6}{( s + 2 ) ^{3}} + \frac{6}{( s + 3 ) ^{4}}

What is laplacetransform (1+te^{-t})^3 ?
The answer to laplacetransform (1+te^{-t})^3 is 1/s+3/((s+1)^2)+6/((s+2)^3)+6/((s+3)^4)