What is inverse oflaplace 3/(s(6s^2+5s+1)) ?

The answer to inverse oflaplace 3/(s(6s^2+5s+1)) is 3\H(t)-9e^{-t/3}+6e^{-t/2}

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inverse oflaplace 3/(s(6s^2+5s+1))

inverse laplace

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

3 H (t) - 9 e^{- \frac{t}{3}} + 6 e^{- \frac{t}{2}}

Solution steps

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

Take the partial fraction of

\frac{3}{s ( 6 s ^{2} + 5 s + 1 )} : \frac{3}{s} - \frac{27}{3 s + 1} + \frac{12}{2 s + 1}

= L^{- 1} {\frac{3}{s} - \frac{27}{3 s + 1} + \frac{12}{2 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{3}{s}} - L^{- 1} {\frac{27}{3 s + 1}} + L^{- 1} {\frac{12}{2 s + 1}}

L^{- 1} {\frac{3}{s}} : 3 H (t)

L^{- 1} {\frac{27}{3 s + 1}} : 9 e^{- \frac{t}{3}}

L^{- 1} {\frac{12}{2 s + 1}} : 6 e^{- \frac{t}{2}}

= 3 H (t) - 9 e^{- \frac{t}{3}} + 6 e^{- \frac{t}{2}}

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

y^{^{''}} + 45 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 0

n = 1 \sum \infty n e^{3 n}

x \to 0 + lim (\frac{6}{x ^{2}})

\int \frac{x}{x ^{2} + 2} d x

What is inverse oflaplace 3/(s(6s^2+5s+1)) ?
The answer to inverse oflaplace 3/(s(6s^2+5s+1)) is 3\H(t)-9e^{-t/3}+6e^{-t/2}