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

The answer to inverse oflaplace (-5)/(s(s^2-5s+4)) is -5/4 \H(t)+5/3 e^t-5/12 e^{4t}

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

inverse laplace

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

- \frac{5}{4} H (t) + \frac{5}{3} e^{t} - \frac{5}{12} e^{4 t}

Solution steps

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

Take the partial fraction of

\frac{- 5}{s ( s ^{2} - 5 s + 4 )} : - \frac{5}{4 s} + \frac{5}{3 ( s - 1 )} - \frac{5}{12 ( s - 4 )}

= L^{- 1} {- \frac{5}{4 s} + \frac{5}{3 ( s - 1 )} - \frac{5}{12 ( 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{5}{4 s}} + L^{- 1} {\frac{5}{3 ( s - 1 )}} - L^{- 1} {\frac{5}{12 ( s - 4 )}}

L^{- 1} {\frac{5}{4 s}} : \frac{5}{4} H (t)

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

L^{- 1} {\frac{5}{12 ( s - 4 )}} : \frac{5}{12} e^{4 t}

= - \frac{5}{4} H (t) + \frac{5}{3} e^{t} - \frac{5}{12} e^{4 t}

y = - 3 x - 2

x^{^{''}} + 6 x^{^{'}} + 7 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\int_{2}^{5} (4 x - 2) d x

(x)^{^{'}}

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

What is inverse oflaplace (-5)/(s(s^2-5s+4)) ?
The answer to inverse oflaplace (-5)/(s(s^2-5s+4)) is -5/4 \H(t)+5/3 e^t-5/12 e^{4t}