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

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

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laplacetransform e^{-1}sin(t)

laplace transform

e^{- 1} sin (t)

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

Solution steps

L {e^{- 1} sin (t)}

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^{- 1} L {sin (t)}

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

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

Refine

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

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

x = y, x = 20 - y

\int \frac{1}{x ( 4 x ) ^{2} - 9} d x

\frac{\partial}{\partial x} (- \frac{1}{10 + 2 x y})

\int sin (\frac{nπ}{2} x) d x

ln (2 x^{2} + 1)

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