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

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

Popular >

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

f (x) = \frac{3 x - 1}{x}

\int \frac{cos ^{5} ( x )}{sin ^{3} ( x )} d x

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

\int \frac{1}{t ^{2} 2 - t ^{2}} d t

\frac{( cos ( x ) )}{cos ^{2} ( \cdot x ) - 1}

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