What is inverse oflaplace (s^2+2s+5)/(s^2(s+1)) ?

The answer to inverse oflaplace (s^2+2s+5)/(s^2(s+1)) is -3\H(t)+5t+4e^{-t}

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inverse oflaplace (s^2+2s+5)/(s^2(s+1))

inverse laplace

\frac{s ^{2} + 2 s + 5}{s ^{2} ( s + 1 )}

- 3 H (t) + 5 t + 4 e^{- t}

Solution steps

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

Take the partial fraction of

\frac{s ^{2} + 2 s + 5}{s ^{2} ( s + 1 )} : - \frac{3}{s} + \frac{5}{s ^{2}} + \frac{4}{s + 1}

= L^{- 1} {- \frac{3}{s} + \frac{5}{s ^{2}} + \frac{4}{s + 1}}

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}{s}} + L^{- 1} {\frac{5}{s ^{2}}} + L^{- 1} {\frac{4}{s + 1}}

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

L^{- 1} {\frac{5}{s ^{2}}} : 5 t

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

= - 3 H (t) + 5 t + 4 e^{- t}

ma c l a u r in e^{i θ}

\frac{d}{d z} (e^{i z})

\frac{d y}{d t} = 1.7 y

d er i v a t i v e f (x) = (x - 5) (x + 1)

\int 4 e^{3 x + e^{3 x}} d x

What is inverse oflaplace (s^2+2s+5)/(s^2(s+1)) ?
The answer to inverse oflaplace (s^2+2s+5)/(s^2(s+1)) is -3\H(t)+5t+4e^{-t}