What is inverse oflaplace (4s+3)/(s^2+109s+990) ?

The answer to inverse oflaplace (4s+3)/(s^2+109s+990) is -37/89 e^{-10t}+393/89 e^{-99t}

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inverse oflaplace (4s+3)/(s^2+109s+990)

inverse laplace

\frac{4 s + 3}{s ^{2} + 109 s + 990}

- \frac{37}{89} e^{- 10 t} + \frac{393}{89} e^{- 99 t}

Solution steps

L^{- 1} {\frac{4 s + 3}{s ^{2} + 109 s + 990}}

Take the partial fraction of

\frac{4 s + 3}{s ^{2} + 109 s + 990} : - \frac{37}{89 ( s + 10 )} + \frac{393}{89 ( s + 99 )}

= L^{- 1} {- \frac{37}{89 ( s + 10 )} + \frac{393}{89 ( s + 99 )}}

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{37}{89 ( s + 10 )}} + L^{- 1} {\frac{393}{89 ( s + 99 )}}

L^{- 1} {\frac{37}{89 ( s + 10 )}} : \frac{37}{89} e^{- 10 t}

L^{- 1} {\frac{393}{89 ( s + 99 )}} : \frac{393}{89} e^{- 99 t}

= - \frac{37}{89} e^{- 10 t} + \frac{393}{89} e^{- 99 t}

(x + 2)^{2} + (y - 3)^{2} = 37, (- 8, 4)

\int cos^{2} (u) sin^{2} (u) d u

\int (x + 1) cos (2 x^{2} + 4 x) d x

\frac{d}{d x} (6 x^{5} e^{6 x})

\frac{d}{d x} (\frac{1 - 2 x}{6 + x})

What is inverse oflaplace (4s+3)/(s^2+109s+990) ?
The answer to inverse oflaplace (4s+3)/(s^2+109s+990) is -37/89 e^{-10t}+393/89 e^{-99t}