What is inverse oflaplace (18)/(s^2+5s+6)+6/s ?

The answer to inverse oflaplace (18)/(s^2+5s+6)+6/s is 18e^{-2t}-18e^{-3t}+6\H(t)

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inverse oflaplace (18)/(s^2+5s+6)+6/s

inverse laplace

\frac{18}{s ^{2} + 5 s + 6} + \frac{6}{s}

18 e^{- 2 t} - 18 e^{- 3 t} + 6 H (t)

Solution steps

L^{- 1} {\frac{18}{s ^{2} + 5 s + 6} + \frac{6}{s}}

Take the partial fraction of

\frac{18}{s ^{2} + 5 s + 6} : \frac{18}{s + 2} - \frac{18}{s + 3}

= L^{- 1} {\frac{18}{s + 2} - \frac{18}{s + 3} + \frac{6}{s}}

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{18}{s + 2}} - L^{- 1} {\frac{18}{s + 3}} + L^{- 1} {\frac{6}{s}}

L^{- 1} {\frac{18}{s + 2}} : 18 e^{- 2 t}

L^{- 1} {\frac{18}{s + 3}} : 18 e^{- 3 t}

L^{- 1} {\frac{6}{s}} : 6 H (t)

= 18 e^{- 2 t} - 18 e^{- 3 t} + 6 H (t)

(4 x + 3) (x^{2} - 1)^{2}

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

x \to e lim (\frac{1}{2} ln (x))

\frac{d}{d x} (sin ((π x)^{9}))

f (x) = 8 x - 10, (4, 22)

What is inverse oflaplace (18)/(s^2+5s+6)+6/s ?
The answer to inverse oflaplace (18)/(s^2+5s+6)+6/s is 18e^{-2t}-18e^{-3t}+6\H(t)