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

The answer to inverse oflaplace s/(s^2+4s-5) is 1/6 e^t+5/6 e^{-5t}

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

inverse laplace

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

\frac{1}{6} e^{t} + \frac{5}{6} e^{- 5 t}

Solution steps

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

Take the partial fraction of

\frac{s}{s ^{2} + 4 s - 5} : \frac{1}{6 ( s - 1 )} + \frac{5}{6 ( s + 5 )}

= L^{- 1} {\frac{1}{6 ( s - 1 )} + \frac{5}{6 ( 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}{6 ( s - 1 )}} + L^{- 1} {\frac{5}{6 ( s + 5 )}}

L^{- 1} {\frac{1}{6 ( s - 1 )}} : \frac{1}{6} e^{t}

L^{- 1} {\frac{5}{6 ( s + 5 )}} : \frac{5}{6} e^{- 5 t}

= \frac{1}{6} e^{t} + \frac{5}{6} e^{- 5 t}

x^{2} (3 x^{2} - x - 14)

\int \frac{x}{6 x ^{2} + 4} d x

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

y = x^{2} - 4 x + 3, x - y - 1 = 0

\int (2 x^{3} + 6 e^{x}) d x

What is inverse oflaplace s/(s^2+4s-5) ?
The answer to inverse oflaplace s/(s^2+4s-5) is 1/6 e^t+5/6 e^{-5t}