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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Refine

e^{4 t} - 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)

\frac{\partial}{\partial x} ((e^{x} - x) sin (y))

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

x \to 1 lim (x (x + 1))

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

\frac{d}{d x} (x^{4} - 3 x^{2} + 4)

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