What is inverse oflaplace ((11s^2-2s+5))/((s-2)(2s-1)(s+1)) ?

The answer to inverse oflaplace ((11s^2-2s+5))/((s-2)(2s-1)(s+1)) is 5e^{2t}-3/2 e^{t/2}+2e^{-t}

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inverse oflaplace ((11s^2-2s+5))/((s-2)(2s-1)(s+1))

inverse laplace

\frac{( 11 s ^{2} - 2 s + 5 )}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )}

5 e^{2 t} - \frac{3}{2} e^{\frac{t}{2}} + 2 e^{- t}

Solution steps

L^{- 1} {\frac{( 11 s ^{2} - 2 s + 5 )}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )}}

Take the partial fraction of

\frac{11 s ^{2} - 2 s + 5}{( s - 2 ) ( 2 s - 1 ) ( s + 1 )} : \frac{5}{s - 2} - \frac{3}{2 s - 1} + \frac{2}{s + 1}

= L^{- 1} {\frac{5}{s - 2} - \frac{3}{2 s - 1} + \frac{2}{s + 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{5}{s - 2}} - L^{- 1} {\frac{3}{2 s - 1}} + L^{- 1} {\frac{2}{s + 1}}

L^{- 1} {\frac{5}{s - 2}} : 5 e^{2 t}

L^{- 1} {\frac{3}{2 s - 1}} : \frac{3}{2} e^{\frac{t}{2}}

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

= 5 e^{2 t} - \frac{3}{2} e^{\frac{t}{2}} + 2 e^{- t}

- 3 t

\frac{d}{d x} (\frac{2 x ^{5} + 3 x}{x ^{4}})

\int (5 x^{4}) d x

\int 3 x + \frac{1}{x} d x

\frac{df}{d s} = 12 s + \frac{5 s}{9 - s ^{2}}

What is inverse oflaplace ((11s^2-2s+5))/((s-2)(2s-1)(s+1)) ?
The answer to inverse oflaplace ((11s^2-2s+5))/((s-2)(2s-1)(s+1)) is 5e^{2t}-3/2 e^{t/2}+2e^{-t}