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

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

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

inverse laplace

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

\frac{2}{9} H (t) - \frac{2}{9} cos (3 t) + \frac{1}{3} sin (3 t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

Refine

\frac{2}{9} H (t) - \frac{2}{9} cos (3 t) + \frac{9}{9} \frac{1}{3} sin (3 t) : \frac{2}{9} H (t) - \frac{2}{9} cos (3 t) + \frac{1}{3} sin (3 t)

= \frac{2}{9} H (t) - \frac{2}{9} cos (3 t) + \frac{1}{3} sin (3 t)

\int 2 x (x^{2} + 6)^{4} d x

6 \cdot e^{5 t} \cdot cos (2 t) - e^{7 t}

\int a e^{3 x} d x

x \to - 2 lim (x^{2} + 2 x + 4)

f (x) = x^{6} - 4 x^{4} + x + 1

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