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

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

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

laplace transform

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

\frac{2}{s ^{3}} + \frac{2}{( s + 1 ) ^{2} + 4}

Solution steps

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

L {t^{2}} : \frac{2}{s ^{3}}

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

= \frac{2}{s ^{3}} + \frac{2}{( s + 1 ) ^{2} + 4}

\int - e^{- 3 x} d x

\int \frac{e ^{2 x}}{x} d x

\int (3 - x^{2}) \cdot 2^{x} d x

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = a, y^{^{'}} (0) = b

\int_{- 1}^{2} x^{4} d x

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