What is laplacetransform cos(-2t) ?

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

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laplacetransform cos(-2t)

laplace transform

cos (- 2 t)

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

Solution steps

L {cos (- 2 t)}

Use Laplace Transform table:

L {cos (a t)} = \frac{s}{s ^{2} + a ^{2}}

L {cos (- 2 t)} = \frac{s}{s ^{2} + ( - 2 ) ^{2}}

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

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

\int 23 x d x

y^{^{'}} = \frac{y}{t}

y^{^{''}} + 4 y = 16 sin (2 t)

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = a

y = 3 x + 4, (7, 5)

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