What is inverse oflaplace (12(s+1))/(s(s^2+9)) ?

The answer to inverse oflaplace (12(s+1))/(s(s^2+9)) is 4/3 \H(t)-4/3 cos(3t)+4sin(3t)

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inverse oflaplace (12(s+1))/(s(s^2+9))

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{12 ( s + 1 )}{s ( s ^{2} + 9 )} : \frac{4}{3 s} + \frac{- 4 s + 36}{3 ( s ^{2} + 9 )}

= L^{- 1} {\frac{4}{3 s} + \frac{- 4 s + 36}{3 ( s ^{2} + 9 )}}

Expand

= L^{- 1} {- \frac{4 s}{3 ( s ^{2} + 9 )} + \frac{4}{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{4}{3} L^{- 1} {\frac{1}{s}} - \frac{4}{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{4}{3} H (t) - \frac{4}{3} cos (3 t) + \frac{36}{3} \cdot \frac{1}{3} sin (3 t)

Refine

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

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

\frac{\partial}{\partial x} (cos (3 y))

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

\int \frac{1}{4} x^{2} d x

\int 2 x (x^{2} + 4)^{2} d x

t - (t - 5) (- (t - 5))

What is inverse oflaplace (12(s+1))/(s(s^2+9)) ?
The answer to inverse oflaplace (12(s+1))/(s(s^2+9)) is 4/3 \H(t)-4/3 cos(3t)+4sin(3t)