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

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

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

inverse laplace

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

- \frac{3}{4} H (t) + \frac{t}{2} + e^{- t} - \frac{1}{4} e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{3}{4 s} + \frac{1}{2 s ^{2}} + \frac{1}{s + 1} - \frac{1}{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{1}{s + 1}} - L^{- 1} {\frac{1}{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{1}{s + 1}} : e^{- t}

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

= - \frac{3}{4} H (t) + \frac{t}{2} + e^{- t} - \frac{1}{4} e^{- 2 t}

\int sin^{6} (t) cos^{3} (t) d t

x \to 8 lim (8 x + ∣ x - 8 ∣)

\int cos^{2} (x) \cdot sin^{2} (x) d x

\frac{\partial}{\partial x} (\frac{2}{x ^{2} + 1})

2 x y^{3} + x y = 18, (6, 1)

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