What is inverse oflaplace (3(s+1)*e^{-2s})/(s(s+1)(2s+1)) ?

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

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inverse oflaplace (3(s+1)*e^{-2s})/(s(s+1)(2s+1))

inverse laplace

\frac{3 ( s + 1 ) \cdot e ^{- 2 s}}{s ( s + 1 ) ( 2 s + 1 )}

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

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 2) L^{- 1} {\frac{3 ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}} (t - 2)

L^{- 1} {\frac{3 ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}} : 3 H (t) - 3 e^{- \frac{t}{2}}

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

H (t - 2) = {01 t < 2 t \geq 2

= {0 3 - 3 e^{- \frac{t - 2}{2}} t < 2 t \geq 2

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

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

\int t - 2 d t

\int \frac{1}{( x ^{2} + r ^{2} ) ^{\frac{3}{2}}} d x

\frac{e ^{- 3 s}}{s + 3}

\int x - csc (x) \cdot co tan (x)

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