What is inverse oflaplace ((12s+8))/((s-1)^5) ?

The answer to inverse oflaplace ((12s+8))/((s-1)^5) is e^t*2t^3+(5e^tt^4)/6

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inverse oflaplace ((12s+8))/((s-1)^5)

inverse laplace

\frac{( 12 s + 8 )}{( s - 1 ) ^{5}}

e^{t} \cdot 2 t^{3} + \frac{5 e ^{t} t ^{4}}{6}

Solution steps

L^{- 1} {\frac{( 12 s + 8 )}{( s - 1 ) ^{5}}}

Take the partial fraction of

\frac{12 s + 8}{( s - 1 ) ^{5}} : \frac{12}{( s - 1 ) ^{4}} + \frac{20}{( s - 1 ) ^{5}}

= L^{- 1} {\frac{12}{( s - 1 ) ^{4}} + \frac{20}{( s - 1 ) ^{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{12}{( s - 1 ) ^{4}}} + L^{- 1} {\frac{20}{( s - 1 ) ^{5}}}

L^{- 1} {\frac{12}{( s - 1 ) ^{4}}} : e^{t} \cdot 2 t^{3}

L^{- 1} {\frac{20}{( s - 1 ) ^{5}}} : \frac{5 e ^{t} t ^{4}}{6}

= e^{t} \cdot 2 t^{3} + \frac{5 e ^{t} t ^{4}}{6}

(1 - x) y^{^{'}} = y^{2}

\frac{- 8}{5 x + 5}

\frac{d}{d x} (tan (2))

e^{2 x} - sin (3 x)

\int 5 x e^{9 x} d x

What is inverse oflaplace ((12s+8))/((s-1)^5) ?
The answer to inverse oflaplace ((12s+8))/((s-1)^5) is e^t*2t^3+(5e^tt^4)/6