What is inverse oflaplace (s+2)/((s-1)(s+3)^2) ?

The answer to inverse oflaplace (s+2)/((s-1)(s+3)^2) is 3/16 e^t-3/16 e^{-3t}+(e^{-3t}t)/4

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inverse oflaplace (s+2)/((s-1)(s+3)^2)

inverse laplace

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

\frac{3}{16} e^{t} - \frac{3}{16} e^{- 3 t} + \frac{e ^{- 3 t} t}{4}

Solution steps

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

Take the partial fraction of

\frac{s + 2}{( s - 1 ) ( s + 3 ) ^{2}} : \frac{3}{16 ( s - 1 )} - \frac{3}{16 ( s + 3 )} + \frac{1}{4 ( s + 3 ) ^{2}}

= L^{- 1} {\frac{3}{16 ( s - 1 )} - \frac{3}{16 ( s + 3 )} + \frac{1}{4 ( s + 3 ) ^{2}}}

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{3}{16 ( s - 1 )}} - L^{- 1} {\frac{3}{16 ( s + 3 )}} + L^{- 1} {\frac{1}{4 ( s + 3 ) ^{2}}}

L^{- 1} {\frac{3}{16 ( s - 1 )}} : \frac{3}{16} e^{t}

L^{- 1} {\frac{3}{16 ( s + 3 )}} : \frac{3}{16} e^{- 3 t}

L^{- 1} {\frac{1}{4 ( s + 3 ) ^{2}}} : \frac{e ^{- 3 t} t}{4}

= \frac{3}{16} e^{t} - \frac{3}{16} e^{- 3 t} + \frac{e ^{- 3 t} t}{4}

What is inverse oflaplace (s+2)/((s-1)(s+3)^2) ?
The answer to inverse oflaplace (s+2)/((s-1)(s+3)^2) is 3/16 e^t-3/16 e^{-3t}+(e^{-3t}t)/4