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

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

Popular >

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

inverse laplace

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

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

Solution steps

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

Expand

= L^{- 1} {\frac{2 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)}

= 2 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)

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

\frac{d}{d x} (- 2 arcsec (x + 1))

\int sin (a \cdot x) cos (a \cdot x) d x

\int 8 e^{- x} d x

\frac{d y}{d x} = \frac{k}{x}

\frac{\partial}{\partial x} (10 e^{x^{7} y})

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