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

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

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

laplace transform

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

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

Solution steps

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

Rewrite using trig identities

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

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

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

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

Refine

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

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

\frac{\partial}{\partial y} (24 y^{2} - 12 x)

\frac{\partial}{\partial x} (7 Y (x))

\frac{- 5}{s ( 6 ( 5 s + 1 ) )}

f (x) = (7 x^{6} + 4 x^{3})^{4}

\frac{\partial}{\partial x} (2 x - y^{2})

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