What is inverse oflaplace (30)/(s(s+5)(s+4)) ?

The answer to inverse oflaplace (30)/(s(s+5)(s+4)) is 3/2 \H(t)+6e^{-5t}-15/2 e^{-4t}

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inverse oflaplace (30)/(s(s+5)(s+4))

inverse laplace

\frac{30}{s ( s + 5 ) ( s + 4 )}

\frac{3}{2} H (t) + 6 e^{- 5 t} - \frac{15}{2} e^{- 4 t}

Solution steps

L^{- 1} {\frac{30}{s ( s + 5 ) ( s + 4 )}}

Take the partial fraction of

\frac{30}{s ( s + 5 ) ( s + 4 )} : \frac{3}{2 s} + \frac{6}{s + 5} - \frac{15}{2 ( s + 4 )}

= L^{- 1} {\frac{3}{2 s} + \frac{6}{s + 5} - \frac{15}{2 ( 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} {\frac{3}{2 s}} + L^{- 1} {\frac{6}{s + 5}} - L^{- 1} {\frac{15}{2 ( s + 4 )}}

L^{- 1} {\frac{3}{2 s}} : \frac{3}{2} H (t)

L^{- 1} {\frac{6}{s + 5}} : 6 e^{- 5 t}

L^{- 1} {\frac{15}{2 ( s + 4 )}} : \frac{15}{2} e^{- 4 t}

= \frac{3}{2} H (t) + 6 e^{- 5 t} - \frac{15}{2} e^{- 4 t}

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

\frac{d y}{d x} = (y e^{x})

\frac{\partial}{\partial y} (sin (2 x + y))

f (x) = 6 x, at x = 9

\int 2 e^{x} x d x

What is inverse oflaplace (30)/(s(s+5)(s+4)) ?
The answer to inverse oflaplace (30)/(s(s+5)(s+4)) is 3/2 \H(t)+6e^{-5t}-15/2 e^{-4t}