What is inverse oflaplace (3s+4)/(s^2+10s+21) ?

The answer to inverse oflaplace (3s+4)/(s^2+10s+21) is -5/4 e^{-3t}+17/4 e^{-7t}

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inverse oflaplace (3s+4)/(s^2+10s+21)

inverse laplace

\frac{3 s + 4}{s ^{2} + 10 s + 21}

- \frac{5}{4} e^{- 3 t} + \frac{17}{4} e^{- 7 t}

Solution steps

L^{- 1} {\frac{3 s + 4}{s ^{2} + 10 s + 21}}

Take the partial fraction of

\frac{3 s + 4}{s ^{2} + 10 s + 21} : - \frac{5}{4 ( s + 3 )} + \frac{17}{4 ( s + 7 )}

= L^{- 1} {- \frac{5}{4 ( s + 3 )} + \frac{17}{4 ( 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{5}{4 ( s + 3 )}} + L^{- 1} {\frac{17}{4 ( s + 7 )}}

L^{- 1} {\frac{5}{4 ( s + 3 )}} : \frac{5}{4} e^{- 3 t}

L^{- 1} {\frac{17}{4 ( s + 7 )}} : \frac{17}{4} e^{- 7 t}

= - \frac{5}{4} e^{- 3 t} + \frac{17}{4} e^{- 7 t}

f^{^{'}} (x) = 16 x

e^{(- 2 t)}

\int - \frac{3}{x ^{4}} d x

\int \frac{x ^{2} + 3 x + 5}{x ^{3} + 8} d x

\frac{d ^{4}}{d x ^{4}} (e^{2 x})

What is inverse oflaplace (3s+4)/(s^2+10s+21) ?
The answer to inverse oflaplace (3s+4)/(s^2+10s+21) is -5/4 e^{-3t}+17/4 e^{-7t}