What is inverse oflaplace (5se^{-2s})/(2s^2+50) ?

The answer to inverse oflaplace (5se^{-2s})/(2s^2+50) is \H(t-2) 5/2 cos(5(t-2))

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inverse oflaplace (5se^{-2s})/(2s^2+50)

inverse laplace

\frac{5 s e ^{- 2 s}}{2 s ^{2} + 50}

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

Solution steps

L^{- 1} {\frac{5 e ^{- 2 s} s}{2 s ^{2} + 50}}

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{5 s}{2 s ^{2} + 50}} (t - 2)

L^{- 1} {\frac{5 s}{2 s ^{2} + 50}} : \frac{5}{2} cos (5 t)

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

\int \frac{4 x ^{2} - 12 x - 19}{( x + 1 ) ^{2} ( x - 2 )} d x

\frac{d y}{d x} (x + 3) = y

\int e^{2 y} d y

\frac{d y}{d x} = (3 x^{2} \cdot (1 + y^{2}))

y = x^{2} + 9

What is inverse oflaplace (5se^{-2s})/(2s^2+50) ?
The answer to inverse oflaplace (5se^{-2s})/(2s^2+50) is \H(t-2) 5/2 cos(5(t-2))