What is laplacetransform 5cos(2t-1) ?

The answer to laplacetransform 5cos(2t-1) is (5(cos(1)s+2sin(1)))/(s^2+4)

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laplacetransform 5cos(2t-1)

laplace transform

5 cos (2 t - 1)

\frac{5 ( cos ( 1 ) s + 2 sin ( 1 ) )}{s ^{2} + 4}

Solution steps

L {5 cos (2 t - 1)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 5 L {cos (2 t - 1)}

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

= 5 \cdot \frac{cos ( 1 ) s + 2 sin ( 1 )}{s ^{2} + 4}

Refine

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

= \frac{5 ( cos ( 1 ) s + 2 sin ( 1 ) )}{s ^{2} + 4}

\frac{d}{d x} (\frac{x ^{2}}{x ^{3} + 1})

\frac{1}{x ^{10}}

\frac{d}{d x} (\frac{4}{x ^{7}} - e^{- 8 x})

\int \frac{x}{( x + 1 ) ( x + 2 )} d x

\frac{\partial}{\partial x} (1.015 E (m, x) b h^{3} m^{- 1} L^{- 4})

What is laplacetransform 5cos(2t-1) ?
The answer to laplacetransform 5cos(2t-1) is (5(cos(1)s+2sin(1)))/(s^2+4)