What is inverse oflaplace 5/((s^3-10*s^2)) ?

The answer to inverse oflaplace 5/((s^3-10*s^2)) is -1/20 \H(t)-t/2+1/20 e^{10t}

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inverse oflaplace 5/((s^3-10*s^2))

inverse laplace

\frac{5}{( s ^{3} - 10 \cdot s ^{2} )}

- \frac{1}{20} H (t) - \frac{t}{2} + \frac{1}{20} e^{10 t}

Solution steps

L^{- 1} {\frac{5}{( s ^{3} - 10 s ^{2} )}}

Take the partial fraction of

\frac{5}{s ^{3} - 10 s ^{2}} : - \frac{1}{20 s} - \frac{1}{2 s ^{2}} + \frac{1}{20 ( s - 10 )}

= L^{- 1} {- \frac{1}{20 s} - \frac{1}{2 s ^{2}} + \frac{1}{20 ( s - 10 )}}

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}{20 s}} - L^{- 1} {\frac{1}{2 s ^{2}}} + L^{- 1} {\frac{1}{20 ( s - 10 )}}

L^{- 1} {\frac{1}{20 s}} : \frac{1}{20} H (t)

L^{- 1} {\frac{1}{2 s ^{2}}} : \frac{t}{2}

L^{- 1} {\frac{1}{20 ( s - 10 )}} : \frac{1}{20} e^{10 t}

= - \frac{1}{20} H (t) - \frac{t}{2} + \frac{1}{20} e^{10 t}

\int \frac{x ^{2} + 3 x + 1}{x ^{4} + 5 x ^{2} + 4} d x

\frac{d}{d x} ((3 x^{2} - 10)^{- 10})

\frac{\partial}{\partial x} (e^{x} x + 2 y e^{x})

x \to - 2 lim (x^{3})

\frac{3 + 8 x}{9 - 2 x}

What is inverse oflaplace 5/((s^3-10*s^2)) ?
The answer to inverse oflaplace 5/((s^3-10*s^2)) is -1/20 \H(t)-t/2+1/20 e^{10t}