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

The answer to inverse oflaplace (s+3)/(s+1(s+2)) is 1/2 \delta(t)+e^{-t}

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inverse oflaplace (s+3)/(s+1(s+2))

inverse laplace

\frac{s + 3}{s + 1 ( s + 2 )}

\frac{1}{2} δ (t) + e^{- t}

Solution steps

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

Take the partial fraction of

\frac{s + 3}{s + 1 ( s + 2 )} : \frac{1}{2} + \frac{1}{s + 1}

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

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

L^{- 1} {\frac{1}{2}} : \frac{1}{2} δ (t)

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

= \frac{1}{2} δ (t) + e^{- t}

\int - 2 sin^{2} (t) d t

\frac{\partial}{\partial x} (4374 x^{\frac{1}{3}} + 2 y)

x \to 1 lim (\frac{e ^{x} - 1}{x})

\frac{\partial}{\partial z} (x ln (y + z))

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

What is inverse oflaplace (s+3)/(s+1(s+2)) ?
The answer to inverse oflaplace (s+3)/(s+1(s+2)) is 1/2 \delta(t)+e^{-t}