What is inverse oflaplace 4/(s^2+7s+12) ?

The answer to inverse oflaplace 4/(s^2+7s+12) is 4e^{-3t}-4e^{-4t}

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inverse oflaplace 4/(s^2+7s+12)

inverse laplace

\frac{4}{s ^{2} + 7 s + 12}

4 e^{- 3 t} - 4 e^{- 4 t}

Solution steps

L^{- 1} {\frac{4}{s ^{2} + 7 s + 12}}

Take the partial fraction of

\frac{4}{s ^{2} + 7 s + 12} : \frac{4}{s + 3} - \frac{4}{s + 4}

= L^{- 1} {\frac{4}{s + 3} - \frac{4}{s + 4}}

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{4}{s + 3}} - L^{- 1} {\frac{4}{s + 4}}

L^{- 1} {\frac{4}{s + 3}} : 4 e^{- 3 t}

L^{- 1} {\frac{4}{s + 4}} : 4 e^{- 4 t}

= 4 e^{- 3 t} - 4 e^{- 4 t}

x \to 0 lim (e^{x} + 1)

a + b e^{θ}

\frac{d}{d x} ((x^{3} + 2 x^{\frac{1}{2}}) (x^{\frac{1}{3}} + 5))

\int \frac{9}{x ( x ^{4} + 4 )} d x

\int x (x^{2} - 4)^{3} d x

What is inverse oflaplace 4/(s^2+7s+12) ?
The answer to inverse oflaplace 4/(s^2+7s+12) is 4e^{-3t}-4e^{-4t}