What is inverse oflaplace (12(s+2))/(s(s^2+9)) ?

The answer to inverse oflaplace (12(s+2))/(s(s^2+9)) is 8/3 \H(t)-8/3 cos(3t)+4sin(3t)

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inverse oflaplace (12(s+2))/(s(s^2+9))

inverse laplace

\frac{12 ( s + 2 )}{s ( s ^{2} + 9 )}

\frac{8}{3} H (t) - \frac{8}{3} cos (3 t) + 4 sin (3 t)

Solution steps

L^{- 1} {\frac{12 ( s + 2 )}{s ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{12 ( s + 2 )}{s ( s ^{2} + 9 )} : \frac{8}{3 s} + \frac{- 8 s + 36}{3 ( s ^{2} + 9 )}

= L^{- 1} {\frac{8}{3 s} + \frac{- 8 s + 36}{3 ( s ^{2} + 9 )}}

Expand

= L^{- 1} {- \frac{8 s}{3 ( s ^{2} + 9 )} + \frac{8}{3 s} + \frac{36}{3 ( s ^{2} + 9 )}}

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)}

= \frac{8}{3} L^{- 1} {\frac{1}{s}} - \frac{8}{3} L^{- 1} {\frac{s}{s ^{2} + 9}} + \frac{36}{3} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= \frac{8}{3} H (t) - \frac{8}{3} cos (3 t) + \frac{36}{3} \cdot \frac{1}{3} sin (3 t)

Refine

\frac{8}{3} H (t) - \frac{8}{3} cos (3 t) + \frac{36}{3} \frac{1}{3} sin (3 t) : \frac{8}{3} H (t) - \frac{8}{3} cos (3 t) + 4 sin (3 t)

= \frac{8}{3} H (t) - \frac{8}{3} cos (3 t) + 4 sin (3 t)

\int_{0}^{36} \frac{1}{36 - x} d x

y = 5 x, x = 5, y = \frac{5}{x}

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

\frac{d}{d x} (- x^{2} + 1.8 x + 2.5)

y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 1

What is inverse oflaplace (12(s+2))/(s(s^2+9)) ?
The answer to inverse oflaplace (12(s+2))/(s(s^2+9)) is 8/3 \H(t)-8/3 cos(3t)+4sin(3t)