What is laplacetransform 0.03(1-cos(2t)) ?

The answer to laplacetransform 0.03(1-cos(2t)) is (0.12)/(s(s^2+4))

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laplacetransform 0.03(1-cos(2t))

laplace transform

0.03 (1 - cos (2 t))

\frac{0.12}{s ( s ^{2} + 4 )}

Solution steps

L {0.03 (- cos (2 t) + 1)}

Expand

0.03 (1 - cos (2 t)) : 0.03 - 0.03 cos (2 t)

= L {0.03 - 0.03 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 {0.03} - 0.03 L {cos (2 t)}

L {0.03} : \frac{0.03}{s}

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

= \frac{0.03}{s} - 0.03 \cdot \frac{s}{s ^{2} + 4}

Simplify

\frac{0.03}{s} - 0.03 \frac{s}{s ^{2} + 4} : \frac{0.12}{s ( s ^{2} + 4 )}

= \frac{0.12}{s ( s ^{2} + 4 )}

\int_{0}^{\infty} 3 e^{2 x} d x

y \to 0 lim (\frac{4 y ^{5} + 5 y ^{3}}{y ^{4} - y})

\int - \frac{x}{1 + x} d x

6 - y = x^{2}, y + 8 = x^{2}, x - y + 4 = 0, x - y - 6 = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

What is laplacetransform 0.03(1-cos(2t)) ?
The answer to laplacetransform 0.03(1-cos(2t)) is (0.12)/(s(s^2+4))