What is inverse oflaplace (s+1)/(s^2-4s) ?

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

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inverse oflaplace (s+1)/(s^2-4s)

inverse laplace

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

- \frac{1}{4} H (t) + \frac{5}{4} e^{4 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{4} H (t) + \frac{5}{4} e^{4 t}

2 \cdot e^{x^{2} - 4 x}

\int 2 x - 5 d x

\frac{d}{d x} (x ln (x^{2} + 4) + 4 arctan (\frac{x}{2}) - 2 x)

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

\int x + 2 y d y

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