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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

f (x) = 2 x^{2} - 3 x - 2, at x = 6

\int_{- \infty}^{- 1} e^{- 4 t} d t

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

f (x) = \frac{( e ^{x} )}{7 x ^{2} + 1}

\int x^{3} 2 x^{4} + 3 d x

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