What is laplacetransform tcos(3t)+2e^{-4t} ?

The answer to laplacetransform tcos(3t)+2e^{-4t} is -(-s^2+9)/((s^2+9)^2)+2/(s+4)

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laplacetransform tcos(3t)+2e^{-4t}

laplace transform

t cos (3 t) + 2 e^{- 4 t}

- \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}} + \frac{2}{s + 4}

Solution steps

L {t cos (3 t) + 2 e^{- 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 {t cos (3 t)} + 2 L {e^{- 4 t}}

L {t cos (3 t)} : - \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}}

L {e^{- 4 t}} : \frac{1}{s + 4}

= - \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}} + 2 \cdot \frac{1}{s + 4}

Refine

- \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}} + 2 \frac{1}{s + 4} : - \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}} + \frac{2}{s + 4}

= - \frac{- s ^{2} + 9}{( s ^{2} + 9 ) ^{2}} + \frac{2}{s + 4}

\int e^{- 3 x} sin (5 x) d x

\frac{d}{d x} (\frac{c ^{9} - x ^{9}}{c ^{9} + x ^{9}})

\int (\frac{1}{x ^{1.2}}) d x

x \to 0 lim ((- ln (x))^{x})

\int \frac{x e ^{- x}}{2} d x

What is laplacetransform tcos(3t)+2e^{-4t} ?
The answer to laplacetransform tcos(3t)+2e^{-4t} is -(-s^2+9)/((s^2+9)^2)+2/(s+4)