What is inverse oflaplace ((3s+1))/(s^2+4s) ?

The answer to inverse oflaplace ((3s+1))/(s^2+4s) is 1/4 \H(t)+11/4 e^{-4t}

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inverse oflaplace ((3s+1))/(s^2+4s)

inverse laplace

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

\frac{1}{4} H (t) + \frac{11}{4} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{3 s + 1}{s ^{2} + 4 s} : \frac{1}{4 s} + \frac{11}{4 ( s + 4 )}

= L^{- 1} {\frac{1}{4 s} + \frac{11}{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{11}{4 ( s + 4 )}}

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

L^{- 1} {\frac{11}{4 ( s + 4 )}} : \frac{11}{4} e^{- 4 t}

= \frac{1}{4} H (t) + \frac{11}{4} e^{- 4 t}

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

- 2 x^{2} + 4 x - 4, at x = 3

y^{^{'}} - 7 y = 0

x \to 0 + lim (csc (x))

f (t) = e^{t} \cdot cos (t)

What is inverse oflaplace ((3s+1))/(s^2+4s) ?
The answer to inverse oflaplace ((3s+1))/(s^2+4s) is 1/4 \H(t)+11/4 e^{-4t}