What is inverse oflaplace e^{-5s}*1/(s^2) ?

The answer to inverse oflaplace e^{-5s}*1/(s^2) is \H(t-5)(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}*1/(s^2)

inverse laplace

e^{- 5 s} \cdot \frac{1}{s ^{2}}

H (t - 5) (t - 5)

Solution steps

L^{- 1} {e^{- 5 s} \frac{1}{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}}} : t

= H (t - 5) (t - 5)

y^{^{'}} - 3 y = - 3 x

\frac{\partial}{\partial x} (5 y cos (- 9 x))

\int_{- \infty}^{- 1} e^{- 26 t} d t

\int x^{2} (\frac{x ^{3}}{18} + 1)^{5} d x

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

What is inverse oflaplace e^{-5s}*1/(s^2) ?
The answer to inverse oflaplace e^{-5s}*1/(s^2) is \H(t-5)(t-5)