What is inverse oflaplace (e^{-5s})/(s^2-4s+13) ?

The answer to inverse oflaplace (e^{-5s})/(s^2-4s+13) is \H(t-5)e^{2(t-5)} 1/3 sin(3(t-5))

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inverse oflaplace (e^{-5s})/(s^2-4s+13)

inverse laplace

\frac{e ^{- 5 s}}{s ^{2} - 4 s + 13}

H (t - 5) e^{2 (t - 5)} \frac{1}{3} sin (3 (t - 5))

Solution steps

L^{- 1} {\frac{e ^{- 5 s}}{s ^{2} - 4 s + 13}}

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 - 5) L^{- 1} {\frac{1}{s ^{2} - 4 s + 13}} (t - 5)

L^{- 1} {\frac{1}{s ^{2} - 4 s + 13}} : e^{2 t} \frac{1}{3} sin (3 t)

= H (t - 5) e^{2 (t - 5)} \frac{1}{3} sin (3 (t - 5))

y^{^{''}} = 2 + x

(1 + x) \frac{d y}{d x} = 4 y

x (x - 2) \frac{d y}{d x} + 2 y = 0

\frac{d}{d x} (4 x^{3} + \frac{1}{x ^{2}})

x \to 0 lim (- x^{8})

What is inverse oflaplace (e^{-5s})/(s^2-4s+13) ?
The answer to inverse oflaplace (e^{-5s})/(s^2-4s+13) is \H(t-5)e^{2(t-5)} 1/3 sin(3(t-5))