What is inverse oflaplace (e^{-7s})/((s^2-25)) ?

The answer to inverse oflaplace (e^{-7s})/((s^2-25)) is \H(t-7) 1/5 sinh(5(t-7))

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inverse oflaplace (e^{-7s})/((s^2-25))

inverse laplace

\frac{e ^{- 7 s}}{( s ^{2} - 25 )}

H (t - 7) \frac{1}{5} sinh (5 (t - 7))

Solution steps

L^{- 1} {\frac{e ^{- 7 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 - 7) L^{- 1} {\frac{1}{s ^{2} - 25}} (t - 7)

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

= H (t - 7) \frac{1}{5} sinh (5 (t - 7))

\frac{d}{d a} (cos (a) \cdot sin (a))

\int 6 sin^{2} (x co) s^{2} x d x

y = 3 x^{3} - x^{2} - 10 x, y = - x^{2} + 2 x

\frac{d}{d x} (\frac{6}{( x - 1 ) ^{2}})

f (x) = 6 cos (x) + 3 sin (2 x)

What is inverse oflaplace (e^{-7s})/((s^2-25)) ?
The answer to inverse oflaplace (e^{-7s})/((s^2-25)) is \H(t-7) 1/5 sinh(5(t-7))