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Popular >

inverse oflaplace (3se^{-s})/(2s^2+8)

Solution

inverse laplace

\frac{3 s e ^{- s}}{2 s ^{2} + 8}

Solution

H (t - 1) \frac{3}{2} cos (2 (t - 1))

Solution steps

L^{- 1} {\frac{3 e ^{- s} s}{2 s ^{2} + 8}}

Apply inverse transform rule:

if

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 - 1) L^{- 1} {\frac{3 s}{2 s ^{2} + 8}} (t - 1)

L^{- 1} {\frac{3 s}{2 s ^{2} + 8}} : \frac{3}{2} cos (2 t)

= H (t - 1) \frac{3}{2} cos (2 (t - 1))

Popular Examples

limit as x approaches 2-of-2/(sqrt(2-x))

x \to 2 - lim (- \frac{2}{2 - x})

-2y^'+(4y)/x =x^4y^3

- 2 y^{^{'}} + \frac{4 y}{x} = x^{4} y^{3}

(\partial)/(\partial x)(x^3-3xy^2)

\frac{\partial}{\partial x} (x^{3} - 3 x y^{2})

derivative of 2/(1-x^2)

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

integral of 1/(2x-1)

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

Frequently Asked Questions (FAQ)

What is inverse oflaplace (3se^{-s})/(2s^2+8) ?
The answer to inverse oflaplace (3se^{-s})/(2s^2+8) is \H(t-1) 3/2 cos(2(t-1))