What is inverse oflaplace (0.32)/((15.5*s+1)*s) ?

The answer to inverse oflaplace (0.32)/((15.5*s+1)*s) is -0.32e^{-0.06451…t}+0.32\H(t)

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inverse oflaplace (0.32)/((15.5s+1)s)

inverse laplace

\frac{0.32}{( 15.5 \cdot s + 1 ) \cdot s}

- 0.32 e^{- 0.06451 \dots t} + 0.32 H (t)

Solution steps

L^{- 1} {\frac{0.32}{s ( 15.5 s + 1 )}}

Take the partial fraction of

\frac{0.32}{( 15.5 s + 1 ) s} : - \frac{4.96}{15.5 s + 1} + \frac{0.32}{s}

= L^{- 1} {- \frac{4.96}{15.5 s + 1} + \frac{0.32}{s}}

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{4.96}{15.5 s + 1}} + L^{- 1} {\frac{0.32}{s}}

L^{- 1} {\frac{4.96}{15.5 s + 1}} : 0.32 e^{- 0.06451 \dots t}

L^{- 1} {\frac{0.32}{s}} : 0.32 H (t)

= - 0.32 e^{- 0.06451 \dots t} + 0.32 H (t)

\frac{1}{( s ^{2} + 167 s + 12874 )}

x \to 3 lim (x + 8)

\frac{d y}{d x} = \frac{1}{3} y (8 - y), y (0) = 1

2 x^{4} y^{^{'}} = k

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

What is inverse oflaplace (0.32)/((15.5*s+1)*s) ?
The answer to inverse oflaplace (0.32)/((15.5*s+1)*s) is -0.32e^{-0.06451…t}+0.32\H(t)