What is inverse oflaplace (20)/((s^2+4)(s-4)) ?

The answer to inverse oflaplace (20)/((s^2+4)(s-4)) is -cos(2t)-2sin(2t)+e^{4t}

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inverse oflaplace (20)/((s^2+4)(s-4))

inverse laplace

\frac{20}{( s ^{2} + 4 ) ( s - 4 )}

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

Solution steps

L^{- 1} {\frac{20}{( s ^{2} + 4 ) ( s - 4 )}}

Take the partial fraction of

\frac{20}{( s ^{2} + 4 ) ( s - 4 )} : \frac{- s - 4}{s ^{2} + 4} + \frac{1}{s - 4}

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

= - cos (2 t) - 4 \cdot \frac{1}{2} sin (2 t) + e^{4 t}

Refine

- cos (2 t) - 4 \frac{1}{2} sin (2 t) + e^{4 t} : - cos (2 t) - 2 sin (2 t) + e^{4 t}

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

\int (a^{2} \cdot x)^{\frac{( 1 - a )}{a}} d x

x \to 2 lim (tan (\frac{π x}{4}))

\frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

x \to \infty lim (\frac{x}{5})

What is inverse oflaplace (20)/((s^2+4)(s-4)) ?
The answer to inverse oflaplace (20)/((s^2+4)(s-4)) is -cos(2t)-2sin(2t)+e^{4t}