What is laplacetransform 5e^{-2(t-1)} ?

The answer to laplacetransform 5e^{-2(t-1)} is 5/(e(s+2))

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laplacetransform 5e^{-2(t-1)}

laplace transform

5 e^{- 2 (t - 1)}

\frac{5}{e ( s + 2 )}

Solution steps

L {5 e^{- 2 (t - 1)}}

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)}

= 5 L {e^{- 2 t - 1}}

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

= 5 \cdot \frac{1}{e ( s + 2 )}

Refine

5 \frac{1}{e ( s + 2 )} : \frac{5}{e ( s + 2 )}

= \frac{5}{e ( s + 2 )}

n = 1 \sum \infty (- 3)^{- n}

\frac{\partial}{\partial x} (4 y)

\frac{\partial}{\partial x} (6 x + 5 y)

\int \frac{x}{( x + 1 ) ^{4}} d x

n = 1 \sum \infty 3 e^{- 9 n}

What is laplacetransform 5e^{-2(t-1)} ?
The answer to laplacetransform 5e^{-2(t-1)} is 5/(e(s+2))