What is inverse oflaplace (7s^3-9s+1)/(3s^5) ?

The answer to inverse oflaplace (7s^3-9s+1)/(3s^5) is (7t)/3-(t^3)/2+(t^4)/(72)

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inverse oflaplace (7s^3-9s+1)/(3s^5)

inverse laplace

\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}

\frac{7 t}{3} - \frac{t ^{3}}{2} + \frac{t ^{4}}{72}

Solution steps

L^{- 1} {\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}}

Expand

= L^{- 1} {\frac{7 s ^{3}}{3 s ^{5}} + \frac{1}{3 s ^{5}} - \frac{9 s}{3 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)}

= \frac{7}{3} L^{- 1} {\frac{s ^{3}}{s ^{5}}} - \frac{9}{3} L^{- 1} {\frac{s}{s ^{5}}} + \frac{1}{3} L^{- 1} {\frac{1}{s ^{5}}}

L^{- 1} {\frac{s ^{3}}{s ^{5}}} : t

L^{- 1} {\frac{s}{s ^{5}}} : \frac{t ^{3}}{6}

L^{- 1} {\frac{1}{s ^{5}}} : \frac{t ^{4}}{24}

= \frac{7}{3} t - \frac{9}{3} \cdot \frac{t ^{3}}{6} + \frac{1}{3} \cdot \frac{t ^{4}}{24}

Refine

\frac{7}{3} t - \frac{9}{3} \frac{t ^{3}}{6} + \frac{1}{3} \frac{t ^{4}}{24} : \frac{7 t}{3} - \frac{t ^{3}}{2} + \frac{t ^{4}}{72}

= \frac{7 t}{3} - \frac{t ^{3}}{2} + \frac{t ^{4}}{72}

What is inverse oflaplace (7s^3-9s+1)/(3s^5) ?
The answer to inverse oflaplace (7s^3-9s+1)/(3s^5) is (7t)/3-(t^3)/2+(t^4)/(72)