What is inverse oflaplace (s-1)/(s(s+2.5)) ?

The answer to inverse oflaplace (s-1)/(s(s+2.5)) is -2/5 \H(t)+1.4e^{-2.5t}

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inverse oflaplace (s-1)/(s(s+2.5))

inverse laplace

\frac{s - 1}{s ( s + 2.5 )}

- \frac{2}{5} H (t) + 1.4 e^{- 2.5 t}

Solution steps

L^{- 1} {\frac{s - 1}{s ( s + 2.5 )}}

Take the partial fraction of

\frac{s - 1}{s ( s + 2.5 )} : - \frac{2}{5 s} + \frac{1.4}{s + 2.5}

= L^{- 1} {- \frac{2}{5 s} + \frac{1.4}{s + 2.5}}

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{2}{5 s}} + L^{- 1} {\frac{1.4}{s + 2.5}}

L^{- 1} {\frac{2}{5 s}} : \frac{2}{5} H (t)

L^{- 1} {\frac{1.4}{s + 2.5}} : 1.4 e^{- 2.5 t}

= - \frac{2}{5} H (t) + 1.4 e^{- 2.5 t}

\frac{\partial}{\partial x} (3 x z e^{4 x y})

x \to - 3 lim (3 x^{2} - 4 x + 1)

(x + 1) (\frac{d y}{d x}) + (2 x - 1) y = e^{- 2 x}

x \to - 1 lim ((x + 2 x^{3})^{5})

f (x) = \frac{1}{2} - x^{2}, at x = - \frac{1}{2}

What is inverse oflaplace (s-1)/(s(s+2.5)) ?
The answer to inverse oflaplace (s-1)/(s(s+2.5)) is -2/5 \H(t)+1.4e^{-2.5t}