What is inverse oflaplace (s^2-3)/(s+1*s+2) ?

The answer to inverse oflaplace (s^2-3)/(s+1*s+2) is (\delta'(t)-\delta(t)-2e^{-t})/2

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inverse oflaplace (s^2-3)/(s+1*s+2)

inverse laplace

\frac{s ^{2} - 3}{s + 1 \cdot s + 2}

\frac{δ ' ( t ) - δ ( t ) - 2 e ^{- t}}{2}

Solution steps

L^{- 1} {\frac{s ^{2} - 3}{s + 1 \cdot s + 2}}

Factor

s + 1 \cdot s + 2 : 2 (s + 1)

= L^{- 1} {\frac{s ^{2} - 3}{2 ( s + 1 )}}

Expand

= L^{- 1} {\frac{s ^{2}}{2 ( s + 1 )} - \frac{3}{2 ( s + 1 )}}

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)}

= \frac{1}{2} L^{- 1} {\frac{s ^{2}}{s + 1}} - \frac{3}{2} L^{- 1} {\frac{1}{s + 1}}

L^{- 1} {\frac{s ^{2}}{s + 1}} : δ^{'} (t) - δ (t) + e^{- t}

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

= \frac{1}{2} (δ^{^{'}} (t) - δ (t) + e^{- t}) - \frac{3}{2} e^{- t}

Simplify

\frac{1}{2} (δ^{'} (t) - δ (t) + e^{- t}) - \frac{3}{2} e^{- t} : \frac{δ ^{'} ( t ) - δ ( t ) - 2 e ^{- t}}{2}

= \frac{δ ^{'} ( t ) - δ ( t ) - 2 e ^{- t}}{2}

x \to 0 lim (x^{5} e^{- x^{4}})

\frac{\partial}{\partial x} (e^{- 8 t} cos (π x))

- x + 5, x^{2}

3 y d x + (x - x y) d y = 0

f (x) = \frac{10}{x}

What is inverse oflaplace (s^2-3)/(s+1*s+2) ?
The answer to inverse oflaplace (s^2-3)/(s+1*s+2) is (\delta'(t)-\delta(t)-2e^{-t})/2