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

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

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

inverse laplace

\frac{2 s + 5}{s ^{2} - 4 s - 5}

- \frac{1}{2} e^{- t} + \frac{5}{2} e^{5 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

L^{- 1} {\frac{5}{2 ( s - 5 )}} : \frac{5}{2} e^{5 t}

= - \frac{1}{2} e^{- t} + \frac{5}{2} e^{5 t}

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

\int sin (x) cos (6 x) d x

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{7})

arctan (x + y) = y^{2} + \frac{π}{4}, (1, 0)

\int (x^{3} + 5 x^{2} - 2) e^{2 x} d x

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