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

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

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

inverse laplace

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

\frac{5}{4} H (t) - 2 e^{t} + \frac{3}{4} e^{4 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= \frac{5}{4} H (t) - 2 e^{t} + \frac{3}{4} e^{4 t}

\frac{d}{d x} (\frac{3}{5})

\frac{1 + 2 x}{1 - 2 x}

n = 9 \sum \infty e^{3 - 6 n}

(y^{2} - 1) d x - (2 y + x y) d y = 0

y = x^{2} - 3 x + 1

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