What is inverse oflaplace 2/(5(s+3)^2+125) ?

The answer to inverse oflaplace 2/(5(s+3)^2+125) is e^{-3t} 2/25 sin(5t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 2/(5(s+3)^2+125)

inverse laplace

\frac{2}{5 ( s + 3 ) ^{2} + 125}

e^{- 3 t} \frac{2}{25} sin (5 t)

Solution steps

L^{- 1} {\frac{2}{5 ( s + 3 ) ^{2} + 125}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{2}{5 ( s + 3 ) ^{2} + 125} : a = - 3, F (s) = \frac{2}{5 s ^{2} + 125}

= e^{- 3 t} L^{- 1} {\frac{2}{5 s ^{2} + 125}}

L^{- 1} {\frac{2}{5 s ^{2} + 125}} : \frac{2}{25} sin (5 t)

= e^{- 3 t} \frac{2}{25} sin (5 t)

\int \frac{lo g _{10} ( lo g _{10} ( x ) )}{x} d x

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

\frac{\partial}{\partial x} (e^{- y} + e^{- x})

f (x) = 4 sin (4 x)

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

What is inverse oflaplace 2/(5(s+3)^2+125) ?
The answer to inverse oflaplace 2/(5(s+3)^2+125) is e^{-3t} 2/25 sin(5t)