What is inverse oflaplace (2s^2+4s+5)/(s(s+1)) ?

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

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inverse oflaplace (2s^2+4s+5)/(s(s+1))

inverse laplace

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

2 δ (t) + 5 H (t) - 3 e^{- t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {2} : 2 δ (t)

L^{- 1} {\frac{5}{s}} : 5 H (t)

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

= 2 δ (t) + 5 H (t) - 3 e^{- t}

\frac{d}{d x} (x^{3} sin (x^{2}))

x^{2} y^{^{'}} + x y = 2

- 2 x^{4}

y = \frac{6}{x}

\int \frac{( 3 x + 1 ) ^{\frac{1}{2}}}{2} d x

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