What is inverse oflaplace (4e^{-2s})/(3s^2+48) ?

The answer to inverse oflaplace (4e^{-2s})/(3s^2+48) is \H(t-2) 1/3 sin(4(t-2))

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inverse oflaplace (4e^{-2s})/(3s^2+48)

inverse laplace

\frac{4 e ^{- 2 s}}{3 s ^{2} + 48}

H (t - 2) \frac{1}{3} sin (4 (t - 2))

Solution steps

L^{- 1} {\frac{4 e ^{- 2 s}}{3 s ^{2} + 48}}

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{4}{3 s ^{2} + 48}} (t - 2)

L^{- 1} {\frac{4}{3 s ^{2} + 48}} : \frac{1}{3} sin (4 t)

= H (t - 2) \frac{1}{3} sin (4 (t - 2))

(2 y^{2} x - 3) d x + (2 y x^{2} + 4) d y = 0

\frac{s + 6}{s ^{2} + 6 s + 25}

\int \frac{t - 3}{t + 1} d t

y = x^{3} - 2 x^{2} + 2, y = 3 x^{2} + 5 x - 23

\frac{( x ^{2} + 1 ) d y}{d x} + 6 x (y - 1) = 0, y (0) = 3

What is inverse oflaplace (4e^{-2s})/(3s^2+48) ?
The answer to inverse oflaplace (4e^{-2s})/(3s^2+48) is \H(t-2) 1/3 sin(4(t-2))