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

The answer to inverse oflaplace 1/((s-1)^2(s+5)) is -1/36 e^t+(e^tt)/6+1/36 e^{-5t}

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

inverse laplace

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

- \frac{1}{36} e^{t} + \frac{e ^{t} t}{6} + \frac{1}{36} e^{- 5 t}

Solution steps

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

Take the partial fraction of

\frac{1}{( s - 1 ) ^{2} ( s + 5 )} : - \frac{1}{36 ( s - 1 )} + \frac{1}{6 ( s - 1 ) ^{2}} + \frac{1}{36 ( s + 5 )}

= L^{- 1} {- \frac{1}{36 ( s - 1 )} + \frac{1}{6 ( s - 1 ) ^{2}} + \frac{1}{36 ( s + 5 )}}

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}{36 ( s - 1 )}} + L^{- 1} {\frac{1}{6 ( s - 1 ) ^{2}}} + L^{- 1} {\frac{1}{36 ( s + 5 )}}

L^{- 1} {\frac{1}{36 ( s - 1 )}} : \frac{1}{36} e^{t}

L^{- 1} {\frac{1}{6 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{6}

L^{- 1} {\frac{1}{36 ( s + 5 )}} : \frac{1}{36} e^{- 5 t}

= - \frac{1}{36} e^{t} + \frac{e ^{t} t}{6} + \frac{1}{36} e^{- 5 t}

f (x) = sin (sin (x)), at x = 4 π

y^{^{'}} = y (1 - y)

y = e^{4 x^{2}} + 1

f (x) = tan (x), at x = 1

19 x, x^{2}

What is inverse oflaplace 1/((s-1)^2(s+5)) ?
The answer to inverse oflaplace 1/((s-1)^2(s+5)) is -1/36 e^t+(e^tt)/6+1/36 e^{-5t}