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

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

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

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

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

= \frac{1}{4} H (t) + \frac{3}{4} e^{- 4 t}

x \to 0 lim (5 x - 3)

f (x) = x^{2} + 3

(x ln (x))^{^{'}}

\frac{d ^{2} y}{d x ^{2}} + \frac{d}{d x} (2 y) + y = 0

\frac{\partial}{\partial x} (1 + 6 x^{2} + 3 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)+3/4 e^{-4t}