What is inverse oflaplace (10)/((s-2)(s^3+s)) ?

The answer to inverse oflaplace (10)/((s-2)(s^3+s)) is e^{2t}-5\H(t)+4cos(t)-2sin(t)

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inverse oflaplace (10)/((s-2)(s^3+s))

inverse laplace

\frac{10}{( s - 2 ) ( s ^{3} + s )}

e^{2 t} - 5 H (t) + 4 cos (t) - 2 sin (t)

Solution steps

L^{- 1} {\frac{10}{( s - 2 ) ( s ^{3} + s )}}

Take the partial fraction of

\frac{10}{( s - 2 ) ( s ^{3} + s )} : \frac{1}{s - 2} - \frac{5}{s} + \frac{4 s - 2}{s ^{2} + 1}

= L^{- 1} {\frac{1}{s - 2} - \frac{5}{s} + \frac{4 s - 2}{s ^{2} + 1}}

Expand

= L^{- 1} {\frac{1}{s - 2} + \frac{4 s}{s ^{2} + 1} - \frac{5}{s} - \frac{2}{s ^{2} + 1}}

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 - 2}} - 5 L^{- 1} {\frac{1}{s}} + 4 L^{- 1} {\frac{s}{s ^{2} + 1}} - 2 L^{- 1} {\frac{1}{s ^{2} + 1}}

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

L^{- 1} {\frac{1}{s}} : H (t)

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

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

= e^{2 t} - 5 H (t) + 4 cos (t) - 2 sin (t)

2 t y^{^{'}} = 4 y

\frac{d P}{d t} = 0.08 P (1 - \frac{P}{1000})

\int \frac{x ^{2}}{( 21 + 4 x - x ^{2} )} d x

\frac{d}{d x} (\frac{\frac{d}{d x} ( e ^{x} )}{1 + x})

\int 2 x (8 x^{2} - 20)^{2} d x

What is inverse oflaplace (10)/((s-2)(s^3+s)) ?
The answer to inverse oflaplace (10)/((s-2)(s^3+s)) is e^{2t}-5\H(t)+4cos(t)-2sin(t)