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

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

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

inverse laplace

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

3 e^{- \frac{t}{2}} - 3 e^{- t}

Solution steps

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

Take the partial fraction of

\frac{3}{2 s ^{2} + 3 s + 1} : \frac{6}{2 s + 1} - \frac{3}{s + 1}

= L^{- 1} {\frac{6}{2 s + 1} - \frac{3}{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{6}{2 s + 1}} - L^{- 1} {\frac{3}{s + 1}}

L^{- 1} {\frac{6}{2 s + 1}} : 3 e^{- \frac{t}{2}}

L^{- 1} {\frac{3}{s + 1}} : 3 e^{- t}

= 3 e^{- \frac{t}{2}} - 3 e^{- t}

\frac{d y}{d x} = 2 y e^{x + 7}

x \to 1 - lim (3)

\frac{x ^{2}}{x + 1}

\frac{d}{d y} (cos (x) sin (y))

\frac{3}{sin ( x ) + cos ( x )}

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