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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

H (t - 1) \frac{4}{3} e^{- 2 (t - 1)} sin (3 (t - 1))

Solution steps

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

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

= H (t - 1) \frac{4}{3} e^{- 2 (t - 1)} sin (3 (t - 1))

\int \frac{1}{x ^{2} - 6 x - 13} d x

y^{^{'}} = 0.0017 y (1500 - y)

\int \frac{sin ( 3 x )}{cos ( 3 x )} d x

\frac{s}{s + 2}

\frac{d}{d x} (e^{x^{2} - 5 x})

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