What is laplacetransform 3+e^{-0.1t}cos(4t) ?

The answer to laplacetransform 3+e^{-0.1t}cos(4t) is 3/s+(s+0.1)/((s+0.1)^2+16)

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laplacetransform 3+e^{-0.1t}cos(4t)

laplace transform

3 + e^{- 0.1 t} cos (4 t)

\frac{3}{s} + \frac{s + 0.1}{( s + 0.1 ) ^{2} + 16}

Solution steps

L {3 + e^{- 0.1 t} cos (4 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 {3} + L {e^{- 0.1 t} cos (4 t)}

L {3} : \frac{3}{s}

L {e^{- 0.1 t} cos (4 t)} : \frac{s + 0.1}{( s + 0.1 ) ^{2} + 16}

= \frac{3}{s} + \frac{s + 0.1}{( s + 0.1 ) ^{2} + 16}

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\frac{d}{d x} (in (x))

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

What is laplacetransform 3+e^{-0.1t}cos(4t) ?
The answer to laplacetransform 3+e^{-0.1t}cos(4t) is 3/s+(s+0.1)/((s+0.1)^2+16)