What is inverse oflaplace ((3s+5))/(2s^2+6s+4) ?

The answer to inverse oflaplace ((3s+5))/(2s^2+6s+4) is e^{-t}+1/2 e^{-2t}

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inverse oflaplace ((3s+5))/(2s^2+6s+4)

inverse laplace

\frac{( 3 s + 5 )}{2 s ^{2} + 6 s + 4}

e^{- t} + \frac{1}{2} e^{- 2 t}

Solution steps

L^{- 1} {\frac{( 3 s + 5 )}{2 s ^{2} + 6 s + 4}}

Take the partial fraction of

\frac{3 s + 5}{2 s ^{2} + 6 s + 4} : \frac{1}{s + 1} + \frac{1}{2 ( s + 2 )}

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

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

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

= e^{- t} + \frac{1}{2} e^{- 2 t}

\int x^{- 2} (x^{6} - 2 x^{3} + 1) d x

\frac{d}{d x} (3 x^{- 2} - \frac{1}{x})

(3 x^{2} y^{2}) d x + (2 x^{3} y + 4 y^{3}) d y = 0

x \to - \frac{π}{3} lim (- 2 cos (x))

\frac{d}{d x} (sin (x^{2}) + cos^{2} (x))

What is inverse oflaplace ((3s+5))/(2s^2+6s+4) ?
The answer to inverse oflaplace ((3s+5))/(2s^2+6s+4) is e^{-t}+1/2 e^{-2t}