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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

H (t - 5) e^{- 2 (t - 5)}

Solution steps

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

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}} (t - 5)

L^{- 1} {\frac{1}{s + 2}} : e^{- 2 t}

= H (t - 5) e^{- 2 (t - 5)}

\int \frac{1}{64 e ^{- 4 x} + e ^{4 x}} d x

\int_{1}^{e} x ln (x) d x

\int \frac{4}{( x ln ( x ) )} d x

y = x^{2} - 2 x - 2, (3, 1)

\int \frac{1}{y ( 3 - y )} d y

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