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}+(9e^{-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{9 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{9}{4 ( s + 4 ) ^{2}}

= L^{- 1} {- \frac{5}{16 s} + \frac{5}{16 ( s + 4 )} + \frac{9}{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{9}{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{9}{4 ( s + 4 ) ^{2}}} : \frac{9 e ^{- 4 t} t}{4}

= - \frac{5}{16} H (t) + \frac{5}{16} e^{- 4 t} + \frac{9 e ^{- 4 t} t}{4}

y = x^{2} - 4 x + 3, x - y - 1 = 0

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

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

\int (- 3 x^{2} + 2 x + 5) d x

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

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}+(9e^{-4t}t)/4