What is inverse oflaplace (-5e^{(-s)})/((s^2+6s+25)) ?

The answer to inverse oflaplace (-5e^{(-s)})/((s^2+6s+25)) is -5/4 e^{-3(t-1)}\H(t-1)sin(4(t-1))

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inverse oflaplace (-5e^{(-s)})/((s^2+6s+25))

inverse laplace

\frac{- 5 e ^{(- s)}}{( s ^{2} + 6 s + 25 )}

- \frac{5}{4} e^{- 3 (t - 1)} H (t - 1) sin (4 (t - 1))

Solution steps

L^{- 1} {\frac{- 5 e ^{(- s)}}{( s ^{2} + 6 s + 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 - 1) L^{- 1} {\frac{- 5}{s ^{2} + 6 s + 25}} (t - 1)

L^{- 1} {\frac{- 5}{s ^{2} + 6 s + 25}} : - \frac{5}{4} e^{- 3 t} sin (4 t)

= H (t - 1) (- \frac{5}{4} e^{- 3 (t - 1)} sin (4 (t - 1)))

= - \frac{5}{4} e^{- 3 (t - 1)} H (t - 1) sin (4 (t - 1))

f (x) = ln (sech (2 x))

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

\frac{d y}{d x} = tan^{2} (x)

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 4}{2 x - 5})

\int \frac{1}{( cos ( x ) ) ^{2}} d x

What is inverse oflaplace (-5e^{(-s)})/((s^2+6s+25)) ?
The answer to inverse oflaplace (-5e^{(-s)})/((s^2+6s+25)) is -5/4 e^{-3(t-1)}\H(t-1)sin(4(t-1))