What is laplacetransform (1-cos(2(3t)))/2 ?

The answer to laplacetransform (1-cos(2(3t)))/2 is (18)/(s(s^2+36))

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laplacetransform (1-cos(2(3t)))/2

laplace transform

\frac{1 - cos ( 2 ( 3 t ) )}{2}

\frac{18}{s ( s ^{2} + 36 )}

Solution steps

L {\frac{1 - cos ( 2 ( 3 t ) )}{2}}

Expand

= L {- \frac{cos ( 2 \cdot 3 t )}{2} + \frac{1}{2}}

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 {\frac{1}{2}} - \frac{1}{2} L {cos (2 \cdot 3 t)}

L {\frac{1}{2}} : \frac{1}{2 s}

L {cos (2 \cdot 3 t)} : \frac{s}{s ^{2} + 36}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 36}

Simplify

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 36} : \frac{18}{s ( s ^{2} + 36 )}

= \frac{18}{s ( s ^{2} + 36 )}

x \to 0 lim (\frac{( 2 ^{x} - 7 ^{x} )}{x})

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

\frac{1}{y ^{2}} - \frac{3}{y ^{4}}

\frac{d y}{d x} = e^{x - y}

f (x) = x^{3} ln (2 x), at x = 1

What is laplacetransform (1-cos(2(3t)))/2 ?
The answer to laplacetransform (1-cos(2(3t)))/2 is (18)/(s(s^2+36))