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

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

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inverse oflaplace ((x+5))/((x^2+5x+6))

inverse laplace

\frac{( x + 5 )}{( x ^{2} + 5 x + 6 )}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e - t

\frac{\partial}{\partial y} (x^{3} + x y^{2})

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 2 e^{x}

\int 2 x + 4 x ln (8 x) d x

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