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

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

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

inverselaplace

cos (3 t) + \frac{2}{3} sin (3 t)

Solution steps

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

Expand

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

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

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

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

= cos (3 t) + 2 \cdot \frac{1}{3} sin (3 t)

Refine

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

= cos (3 t) + \frac{2}{3} sin (3 t)

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