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

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

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

inverse laplace

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

2 e^{- t} - e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{2}{s + 1}} : 2 e^{- t}

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

= 2 e^{- t} - e^{- 2 t}

\frac{\partial}{\partial x} (- y cos (x))

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- x}

\frac{\partial}{\partial x} (x \cdot sin (x + y))

\int (1 - sin (x)) d x

y = 6 tan (x), at x = \frac{π}{3}

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