What is laplacetransform (te)^3e^{-2(t+1)} ?

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

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

laplace transform

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

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

Solution steps

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

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

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

L {t^{3} e^{- 2 t + 1}} : \frac{6 e}{( s + 2 ) ^{4}}

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

Refine

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

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

n = 5 \sum \infty e^{5 - 2 n}

\frac{3}{s ^{2} - 2 s - 15}

\int (\frac{7}{x ^{2} + 1}) d x

\frac{d}{d x} (csc^{3} (x^{2}))

\frac{d}{d x} ((1 + x^{3})^{2020})

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