What is inverse oflaplace (3se^{-2s})/(5s^2+45) ?

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

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inverse oflaplace (3se^{-2s})/(5s^2+45)

inverse laplace

\frac{3 s e ^{- 2 s}}{5 s ^{2} + 45}

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

Solution steps

L^{- 1} {\frac{3 e ^{- 2 s} s}{5 s ^{2} + 45}}

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}{5 s ^{2} + 45}} (t - 2)

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

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

\int 4 t^{4} e^{2 t^{5}} d t

y^{^{''}} + 4 y = cos (2 x)

\int x^{2} arctan (3 x) d x

\frac{\partial}{\partial x} (4 x + 5)

f (x) = x^{2} - 3

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