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

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

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

laplace transform

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

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

Solution steps

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

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} - L {e^{- t} sin (2 t)}

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

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

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

\frac{d}{d x} (8 x (x^{2} + 3)^{3})

f (x) = 16 - x^{2}

(4 x)^{^{'}}

\int (5 x^{2} + 5 x + 3)^{4} (2 x + 1) d x

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

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