What is inverse oflaplace ((3se^{(-2s)}))/(4s^2+36) ?

The answer to inverse oflaplace ((3se^{(-2s)}))/(4s^2+36) is \H(t-2) 3/4 cos(3(t-2))

inverse oflaplace ((3se^{(-2s)}))/(4s^2+36)

inverse laplace

\frac{( 3 s e ^{(- 2 s)} )}{4 s ^{2} + 36}

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

Solution steps

L^{- 1} {\frac{( 3 e ^{(- 2 s)} s )}{4 s ^{2} + 36}}

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}{4 s ^{2} + 36}} (t - 2)

L^{- 1} {\frac{3 s}{4 s ^{2} + 36}} : \frac{3}{4} cos (3 t)

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

\int x^{2} cos (x) - 2 x cos (x) d x

x \to 9 lim (\frac{x - 9}{∣ x - 9 ∣})

x \to 0 lim (ln (x))

in v erse l a pl a ce \frac{5}{( s ^{3} - 10 \cdot s ^{2} )}

\int \frac{x ^{2} + 3 x + 1}{x ^{4} + 5 x ^{2} + 4} d x

What is inverse oflaplace ((3se^{(-2s)}))/(4s^2+36) ?
The answer to inverse oflaplace ((3se^{(-2s)}))/(4s^2+36) is \H(t-2) 3/4 cos(3(t-2))