What is inverse oflaplace (1+s)/(s^3-2s^2) ?

The answer to inverse oflaplace (1+s)/(s^3-2s^2) is -3/4 \H(t)-t/2+3/4 e^{2t}

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inverse oflaplace (1+s)/(s^3-2s^2)

inverse laplace

\frac{1 + s}{s ^{3} - 2 s ^{2}}

- \frac{3}{4} H (t) - \frac{t}{2} + \frac{3}{4} e^{2 t}

Solution steps

L^{- 1} {\frac{1 + s}{s ^{3} - 2 s ^{2}}}

Take the partial fraction of

\frac{1 + s}{s ^{3} - 2 s ^{2}} : - \frac{3}{4 s} - \frac{1}{2 s ^{2}} + \frac{3}{4 ( s - 2 )}

= L^{- 1} {- \frac{3}{4 s} - \frac{1}{2 s ^{2}} + \frac{3}{4 ( s - 2 )}}

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{3}{4 s}} - L^{- 1} {\frac{1}{2 s ^{2}}} + L^{- 1} {\frac{3}{4 ( s - 2 )}}

L^{- 1} {\frac{3}{4 s}} : \frac{3}{4} H (t)

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

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

= - \frac{3}{4} H (t) - \frac{t}{2} + \frac{3}{4} e^{2 t}

What is inverse oflaplace (1+s)/(s^3-2s^2) ?
The answer to inverse oflaplace (1+s)/(s^3-2s^2) is -3/4 \H(t)-t/2+3/4 e^{2t}