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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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}

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

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

x \to 0 - lim (\frac{3}{1 + e ^{\frac{1}{x}}})

\int x (7 - x)^{2} d x

1 + x

\int 6 x - 5 d x

\int 5 x^{4} d x

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