What is laplacetransform 2cos^2(t) ?

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

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laplacetransform 2cos^2(t)

laplace transform

2 cos^{2} (t)

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

Solution steps

L {2 cos^{2} (t)}

Rewrite using trig identities

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}

y = ln (\frac{x + 3}{x ^{2}})

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

n = 1 \sum \infty \frac{n !}{3 n ! - 1}

y = 3 x + 1, y = 13 + 2 x - x^{2}

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