What is inverse oflaplace 5/((s+4)^2(s+1)) ?

The answer to inverse oflaplace 5/((s+4)^2(s+1)) is -5/9 e^{-4t}-(5e^{-4t}t)/3+5/9 e^{-t}

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inverse oflaplace 5/((s+4)^2(s+1))

inverse laplace

\frac{5}{( s + 4 ) ^{2} ( s + 1 )}

- \frac{5}{9} e^{- 4 t} - \frac{5 e ^{- 4 t} t}{3} + \frac{5}{9} e^{- t}

Solution steps

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

Take the partial fraction of

\frac{5}{( s + 4 ) ^{2} ( s + 1 )} : - \frac{5}{9 ( s + 4 )} - \frac{5}{3 ( s + 4 ) ^{2}} + \frac{5}{9 ( s + 1 )}

= L^{- 1} {- \frac{5}{9 ( s + 4 )} - \frac{5}{3 ( s + 4 ) ^{2}} + \frac{5}{9 ( 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}{9 ( s + 4 )}} - L^{- 1} {\frac{5}{3 ( s + 4 ) ^{2}}} + L^{- 1} {\frac{5}{9 ( s + 1 )}}

L^{- 1} {\frac{5}{9 ( s + 4 )}} : \frac{5}{9} e^{- 4 t}

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

L^{- 1} {\frac{5}{9 ( s + 1 )}} : \frac{5}{9} e^{- t}

= - \frac{5}{9} e^{- 4 t} - \frac{5 e ^{- 4 t} t}{3} + \frac{5}{9} e^{- t}

\frac{t}{( t - 1 ) ^{2}}

(1, 1), (- 3, - 3)

y = x^{2} + x - 1.0 E - 13

\int \frac{1}{19 - x ^{2}} d x

3 t^{2} + 13 t

What is inverse oflaplace 5/((s+4)^2(s+1)) ?
The answer to inverse oflaplace 5/((s+4)^2(s+1)) is -5/9 e^{-4t}-(5e^{-4t}t)/3+5/9 e^{-t}