What is inverse oflaplace ((2s+1))/(s^2+5s+6) ?

The answer to inverse oflaplace ((2s+1))/(s^2+5s+6) is -3e^{-2t}+5e^{-3t}

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inverse oflaplace ((2s+1))/(s^2+5s+6)

inverse laplace

\frac{( 2 s + 1 )}{s ^{2} + 5 s + 6}

- 3 e^{- 2 t} + 5 e^{- 3 t}

Solution steps

L^{- 1} {\frac{( 2 s + 1 )}{s ^{2} + 5 s + 6}}

Take the partial fraction of

\frac{2 s + 1}{s ^{2} + 5 s + 6} : - \frac{3}{s + 2} + \frac{5}{s + 3}

= L^{- 1} {- \frac{3}{s + 2} + \frac{5}{s + 3}}

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{3}{s + 2}} + L^{- 1} {\frac{5}{s + 3}}

L^{- 1} {\frac{3}{s + 2}} : 3 e^{- 2 t}

L^{- 1} {\frac{5}{s + 3}} : 5 e^{- 3 t}

= - 3 e^{- 2 t} + 5 e^{- 3 t}

\frac{d y}{d t} = \frac{9}{40} - \frac{y}{200}

t a y l or x^{\frac{11}{2}}, 5

\frac{d}{d x} (5 x^{2} e^{3 x - 7})

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}

\int_{4}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

What is inverse oflaplace ((2s+1))/(s^2+5s+6) ?
The answer to inverse oflaplace ((2s+1))/(s^2+5s+6) is -3e^{-2t}+5e^{-3t}