What is laplacetransform 1-cos(2t) ?

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

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laplacetransform 1-cos(2t)

laplace transform

1 - cos (2 t)

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

Solution steps

L {1 - 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)}

= L {1} - L {cos (2 t)}

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

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

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

Simplify

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

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

roo t s x sin (y)

f^{^{'}} (x) = \frac{x}{8} - 1

s l o p e y = 2 - x^{3}, P (1, 1)

im pl i c i t \frac{d y}{d x}, e \frac{x}{y} = 8 x - y

\frac{d}{d x} (3 arcsin (x^{3}))

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