What is laplacetransform e^{-2t}-3cos(2t) ?

The answer to laplacetransform e^{-2t}-3cos(2t) is 1/(s+2)-(3s)/(s^2+4)

laplacetransform e^{-2t}-3cos(2t)

laplace transform

e^{- 2 t} - 3 cos (2 t)

\frac{1}{s + 2} - \frac{3 s}{s ^{2} + 4}

Solution steps

L {e^{- 2 t} - 3 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 {e^{- 2 t}} - 3 L {cos (2 t)}

L {e^{- 2 t}} : \frac{1}{s + 2}

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

= \frac{1}{s + 2} - 3 \cdot \frac{s}{s ^{2} + 4}

Refine

\frac{1}{s + 2} - 3 \frac{s}{s ^{2} + 4} : \frac{1}{s + 2} - \frac{3 s}{s ^{2} + 4}

= \frac{1}{s + 2} - \frac{3 s}{s ^{2} + 4}

\frac{d}{d x} (- 2 cos (x))

\frac{d}{d x} (1 + tan (x))

d er i v a t i v e f (X) = 4 X^{3} - 1

\int t^{n} d t

\int \frac{1}{( 3 - 5 x ) ^{4}} d x

What is laplacetransform e^{-2t}-3cos(2t) ?
The answer to laplacetransform e^{-2t}-3cos(2t) is 1/(s+2)-(3s)/(s^2+4)