What is inverse oflaplace ((s+5))/(s^2+5s+4) ?

The answer to inverse oflaplace ((s+5))/(s^2+5s+4) is 4/3 e^{-t}-1/3 e^{-4t}

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inverse oflaplace ((s+5))/(s^2+5s+4)

inverse laplace

\frac{( s + 5 )}{s ^{2} + 5 s + 4}

\frac{4}{3} e^{- t} - \frac{1}{3} e^{- 4 t}

Solution steps

L^{- 1} {\frac{( s + 5 )}{s ^{2} + 5 s + 4}}

Take the partial fraction of

\frac{s + 5}{s ^{2} + 5 s + 4} : \frac{4}{3 ( s + 1 )} - \frac{1}{3 ( s + 4 )}

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

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

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

= \frac{4}{3} e^{- t} - \frac{1}{3} e^{- 4 t}

\frac{d}{d x} (x 2 x + 5)

\int \frac{x ^{2} + x ^{- 2}}{x ^{3} - 3 x ^{- 1}} d x

y^{^{'}} + 7 (tan (7 x)) y = 5 cos (7 x)

\frac{x ^{5}}{x ^{3}}

\int \frac{5}{x ^{2} x ^{2} - 9} d x

What is inverse oflaplace ((s+5))/(s^2+5s+4) ?
The answer to inverse oflaplace ((s+5))/(s^2+5s+4) is 4/3 e^{-t}-1/3 e^{-4t}