What is inverse oflaplace (s^2+9)/(s(s+4)) ?

The answer to inverse oflaplace (s^2+9)/(s(s+4)) is \delta(t)+9/4 \H(t)-25/4 e^{-4t}

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inverse oflaplace (s^2+9)/(s(s+4))

inverse laplace

\frac{s ^{2} + 9}{s ( s + 4 )}

δ (t) + \frac{9}{4} H (t) - \frac{25}{4} e^{- 4 t}

Solution steps

L^{- 1} {\frac{s ^{2} + 9}{s ( s + 4 )}}

Take the partial fraction of

\frac{s ^{2} + 9}{s ( s + 4 )} : 1 + \frac{9}{4 s} - \frac{25}{4 ( s + 4 )}

= L^{- 1} {1 + \frac{9}{4 s} - \frac{25}{4 ( s + 4 )}}

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} {1} + L^{- 1} {\frac{9}{4 s}} - L^{- 1} {\frac{25}{4 ( s + 4 )}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{9}{4 s}} : \frac{9}{4} H (t)

L^{- 1} {\frac{25}{4 ( s + 4 )}} : \frac{25}{4} e^{- 4 t}

= δ (t) + \frac{9}{4} H (t) - \frac{25}{4} e^{- 4 t}

\frac{d y}{d x} = \frac{x - 1}{x + 3} \frac{y - 2}{y - 3}

\frac{2 s + 3}{s ^{2} + 6 s + 13}

\frac{x}{x - 2}

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

\frac{1}{5} \frac{d y}{d x} = (\frac{1}{5})^{2} - \frac{1}{5}

What is inverse oflaplace (s^2+9)/(s(s+4)) ?
The answer to inverse oflaplace (s^2+9)/(s(s+4)) is \delta(t)+9/4 \H(t)-25/4 e^{-4t}