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Popular >

inverse oflaplace (2se^{-2s})/(5s^2+45)

Solution

inverse laplace

\frac{2 s e ^{- 2 s}}{5 s ^{2} + 45}

Solution

H (t - 2) \frac{2}{5} cos (3 (t - 2))

Solution steps

L^{- 1} {\frac{2 e ^{- 2 s} s}{5 s ^{2} + 45}}

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 - 2) L^{- 1} {\frac{2 s}{5 s ^{2} + 45}} (t - 2)

L^{- 1} {\frac{2 s}{5 s ^{2} + 45}} : \frac{2}{5} cos (3 t)

= H (t - 2) \frac{2}{5} cos (3 (t - 2))

Popular Examples

(dy)/(dx)= y/((1+x^2))

\frac{d y}{d x} = \frac{y}{( 1 + x ^{2} )}

y^{''}-y^'-2y=4e^{-t}

y^{^{''}} - y^{^{'}} - 2 y = 4 e^{- t}

derivative of f(x)= 2/(x-7)derivative of

f (x) = \frac{2}{x - 7}

integral of 8xe^{-x^2}

\int 8 x e^{- x^{2}} d x

limit as x approaches 2 of (x+3)/(x-2)

x \to 2 lim (\frac{x + 3}{x - 2})

Frequently Asked Questions (FAQ)

What is inverse oflaplace (2se^{-2s})/(5s^2+45) ?
The answer to inverse oflaplace (2se^{-2s})/(5s^2+45) is \H(t-2) 2/5 cos(3(t-2))