What is inverse oflaplace (s*e^{-3s})/(s^2+25) ?

The answer to inverse oflaplace (s*e^{-3s})/(s^2+25) is \H(t-3)cos(5(t-3))

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inverse oflaplace (s*e^{-3s})/(s^2+25)

inverse laplace

\frac{s \cdot e ^{- 3 s}}{s ^{2} + 25}

H (t - 3) cos (5 (t - 3))

Solution steps

L^{- 1} {\frac{e ^{- 3 s} s}{s ^{2} + 25}}

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 - 3) L^{- 1} {\frac{s}{s ^{2} + 25}} (t - 3)

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

= H (t - 3) cos (5 (t - 3))

y^{^{''}} - 10 y^{^{'}} + 74 y = 0

\frac{e ^{8 x}}{x}

n = 1 \sum \infty 3 (\frac{1}{3})^{n - 1}

\frac{d ^{3}}{d x ^{3}} (x^{4})

\frac{\partial}{\partial x} (x^{4} e y^{z})

What is inverse oflaplace (s*e^{-3s})/(s^2+25) ?
The answer to inverse oflaplace (s*e^{-3s})/(s^2+25) is \H(t-3)cos(5(t-3))