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

The answer to inverse oflaplace ((s+5))/(s(s+4)^2) is 5/16 \H(t)-5/16 e^{-4t}-(e^{-4t}t)/4

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

inverse laplace

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

\frac{5}{16} H (t) - \frac{5}{16} e^{- 4 t} - \frac{e ^{- 4 t} t}{4}

Solution steps

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

Take the partial fraction of

\frac{s + 5}{s ( s + 4 ) ^{2}} : \frac{5}{16 s} - \frac{5}{16 ( s + 4 )} - \frac{1}{4 ( s + 4 ) ^{2}}

= L^{- 1} {\frac{5}{16 s} - \frac{5}{16 ( s + 4 )} - \frac{1}{4 ( s + 4 ) ^{2}}}

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}{16 s}} - L^{- 1} {\frac{5}{16 ( s + 4 )}} - L^{- 1} {\frac{1}{4 ( s + 4 ) ^{2}}}

L^{- 1} {\frac{5}{16 s}} : \frac{5}{16} H (t)

L^{- 1} {\frac{5}{16 ( s + 4 )}} : \frac{5}{16} e^{- 4 t}

L^{- 1} {\frac{1}{4 ( s + 4 ) ^{2}}} : \frac{e ^{- 4 t} t}{4}

= \frac{5}{16} H (t) - \frac{5}{16} e^{- 4 t} - \frac{e ^{- 4 t} t}{4}

\int \frac{cos ( x )}{5 + sin ^{2} ( x )} d x

\frac{d y}{d x} = 6 e^{x - y}

\frac{d}{d x} (x^{3} + x)

n = 1 \sum \infty \frac{1}{n ^{2} + 5}

\int \frac{1}{( 4 + x ^{2} )} d x

What is inverse oflaplace ((s+5))/(s(s+4)^2) ?
The answer to inverse oflaplace ((s+5))/(s(s+4)^2) is 5/16 \H(t)-5/16 e^{-4t}-(e^{-4t}t)/4