What is inverse oflaplace (40s+20)/(s^2+9s+14) ?

The answer to inverse oflaplace (40s+20)/(s^2+9s+14) is -12e^{-2t}+52e^{-7t}

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inverse oflaplace (40s+20)/(s^2+9s+14)

inverse laplace

\frac{40 s + 20}{s ^{2} + 9 s + 14}

- 12 e^{- 2 t} + 52 e^{- 7 t}

Solution steps

L^{- 1} {\frac{40 s + 20}{s ^{2} + 9 s + 14}}

Take the partial fraction of

\frac{40 s + 20}{s ^{2} + 9 s + 14} : - \frac{12}{s + 2} + \frac{52}{s + 7}

= L^{- 1} {- \frac{12}{s + 2} + \frac{52}{s + 7}}

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{12}{s + 2}} + L^{- 1} {\frac{52}{s + 7}}

L^{- 1} {\frac{12}{s + 2}} : 12 e^{- 2 t}

L^{- 1} {\frac{52}{s + 7}} : 52 e^{- 7 t}

= - 12 e^{- 2 t} + 52 e^{- 7 t}

\frac{\partial}{\partial x} (x - a)

\int (3 sin (t) - 2 cos (t)) d t

\int \frac{9 - x ^{2}}{x ^{2}} d x

\frac{\partial}{\partial x} (b lo g_{10} (x))

\frac{\partial}{\partial x} ((x_{1} + x_{2})^{2})

What is inverse oflaplace (40s+20)/(s^2+9s+14) ?
The answer to inverse oflaplace (40s+20)/(s^2+9s+14) is -12e^{-2t}+52e^{-7t}