What is laplacetransform e^{-2((t-1))}*1((t-1)) ?

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

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laplacetransform e^{-2((t-1))}*1((t-1))

laplace transform

e^{- 2 ((t - 1))} \cdot 1 ((t - 1))

\frac{e ^{2}}{( s + 2 ) ^{2}} - \frac{e ^{2}}{s + 2}

Solution steps

L {1 \cdot e^{- 2 ((t - 1))} ((t - 1))}

Expand

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

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

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)}

= e^{2} L {e^{- 2 t} t} - e^{2} L {e^{- 2 t}}

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

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

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

Refine

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

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

\int_{0}^{7} e^{x} sin (x) d x

\frac{d}{d x} (2 x^{3} - 48 x^{2})

\int 2 x - 1 d x

\int 16 θ sec^{2} (θ) d θ

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = 0

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