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

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

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

laplace transform

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

\frac{2}{( s + 1 ) ( s ^{2} + 2 s + 5 )}

Solution steps

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

Rewrite using trig identities

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

= \frac{1}{2} L {e^{- t}} - \frac{1}{2} L {e^{- t} cos (2 t)}

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

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

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

Simplify

\frac{1}{2} \frac{1}{s + 1} - \frac{1}{2} \frac{s + 1}{( s + 1 ) ^{2} + 4} : \frac{2}{( s + 1 ) ( s ^{2} + 2 s + 5 )}

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

x y^{^{'''}} - \frac{2}{x} y^{^{'}} = 0

\frac{d}{d x} (sin (p (x)) i x)

cos (8 x)

\frac{s}{( s - 3 ) ^{2}}

\int (x^{2} + x)^{13} (2 x + 1) d x

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