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

The answer to inverse oflaplace ((s^2+1))/(s+5) is \delta'(t)-5\delta(t)+26e^{-5t}

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

inverse laplace

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

δ' (t) - 5 δ (t) + 26 e^{- 5 t}

Solution steps

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

Take the partial fraction of

\frac{s ^{2} + 1}{s + 5} : s - 5 + \frac{26}{s + 5}

= L^{- 1} {s - 5 + \frac{26}{s + 5}}

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} {s} - L^{- 1} {5} + L^{- 1} {\frac{26}{s + 5}}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {5} : 5 δ (t)

L^{- 1} {\frac{26}{s + 5}} : 26 e^{- 5 t}

= δ^{^{'}} (t) - 5 δ (t) + 26 e^{- 5 t}

\int_{0}^{4} (2 x + 4)^{- \frac{1}{2}} d x

\frac{\partial}{\partial y} (x sin (y) sin (z))

\frac{d}{d x} (ln (\frac{x + 6}{x - 6}))

y^{^{'}} = - \frac{2 x y}{1 + x ^{2}}

f (z) = \frac{1}{z}

What is inverse oflaplace ((s^2+1))/(s+5) ?
The answer to inverse oflaplace ((s^2+1))/(s+5) is \delta'(t)-5\delta(t)+26e^{-5t}