What is inverse oflaplace 5/(s^2+3s+2) ?

The answer to inverse oflaplace 5/(s^2+3s+2) is 5e^{-t}-5e^{-2t}

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inverse oflaplace 5/(s^2+3s+2)

inverse laplace

\frac{5}{s ^{2} + 3 s + 2}

5 e^{- t} - 5 e^{- 2 t}

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 3 s + 2}}

Take the partial fraction of

\frac{5}{s ^{2} + 3 s + 2} : \frac{5}{s + 1} - \frac{5}{s + 2}

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

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

L^{- 1} {\frac{5}{s + 2}} : 5 e^{- 2 t}

= 5 e^{- t} - 5 e^{- 2 t}

x \to - 5 lim (\frac{2 x ^{2} + 7 x - 15}{x ^{2} - 25})

\frac{d y}{d x} = \frac{4 sec ( y )}{( x + 3 ) ^{2}}

\int \frac{1}{4 x ^{2} + 9} d x

f (x) = 8 x

\int \frac{1}{9 x ^{2} - 1} d x

What is inverse oflaplace 5/(s^2+3s+2) ?
The answer to inverse oflaplace 5/(s^2+3s+2) is 5e^{-t}-5e^{-2t}