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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

H (t - 2) \frac{5}{2} sin (2 (t - 2))

Solution steps

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

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

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

= H (t - 2) \frac{5}{2} sin (2 (t - 2))

\int (2 x^{3} + 1)^{7} x^{2} d x

\frac{\partial}{\partial y} (T x^{5} y^{4} z^{8})

\frac{x ^{2}}{x ^{3} - 3 x}, \frac{1}{x ^{3} - 3 x}, 2, 11

\frac{d}{d x} (\frac{x e ^{x}}{sin ( x ) + 1})

tan^{2} (3 x)

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