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

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

Popular >

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

inverse laplace

\frac{s + 3}{s ^{2} + 3 s + 2}

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

Solution steps

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

Take the partial fraction of

\frac{s + 3}{s ^{2} + 3 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} (x y - z)

f (x) = \frac{x ^{2} + 6}{9 x - 2}

\frac{\partial}{\partial x} (t^{2})

\int \frac{x ^{2}}{4 - x ^{2}} d x

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

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