What is inverse oflaplace (3e^{-2s})/(4s^2+100) ?

The answer to inverse oflaplace (3e^{-2s})/(4s^2+100) is \H(t-2) 3/20 sin(5(t-2))

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inverse oflaplace (3e^{-2s})/(4s^2+100)

inverse laplace

\frac{3 e ^{- 2 s}}{4 s ^{2} + 100}

H (t - 2) \frac{3}{20} sin (5 (t - 2))

Solution steps

L^{- 1} {\frac{3 e ^{- 2 s}}{4 s ^{2} + 100}}

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}{4 s ^{2} + 100}} (t - 2)

L^{- 1} {\frac{3}{4 s ^{2} + 100}} : \frac{3}{20} sin (5 t)

= H (t - 2) \frac{3}{20} sin (5 (t - 2))

\int \frac{arctan ( 9 x )}{1 + 81 x ^{2}} d x

\frac{d y}{d x} - 2 y = x

\frac{d}{d x} y + y = 4

\int x^{2} n^{2} x d x

\int_{0}^{1} \frac{t ^{2}}{1 + t ^{6}} d t

What is inverse oflaplace (3e^{-2s})/(4s^2+100) ?
The answer to inverse oflaplace (3e^{-2s})/(4s^2+100) is \H(t-2) 3/20 sin(5(t-2))