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}

\frac{3}{x ^{3}} - \frac{1}{x ^{4}}

(26 \frac{2}{3}, 9), (5, 2.5)

x \to 0 lim (x^{2} cos (x))

f (x) = \frac{5}{x}, at x = 5

\int \frac{3 x}{x + 1} 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