What is inverse oflaplace 5/(2(s-1)^2+8) ?

The answer to inverse oflaplace 5/(2(s-1)^2+8) is e^t 5/4 sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 5/(2(s-1)^2+8)

inverse laplace

\frac{5}{2 ( s - 1 ) ^{2} + 8}

e^{t} \frac{5}{4} sin (2 t)

Solution steps

L^{- 1} {\frac{5}{2 ( s - 1 ) ^{2} + 8}}

Apply inverse transform rule:

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

then

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

For

\frac{5}{2 ( s - 1 ) ^{2} + 8} : a = 1, F (s) = \frac{5}{2 s ^{2} + 8}

= e^{t} L^{- 1} {\frac{5}{2 s ^{2} + 8}}

L^{- 1} {\frac{5}{2 s ^{2} + 8}} : \frac{5}{4} sin (2 t)

= e^{t} \frac{5}{4} sin (2 t)

t \to 2 lim (ln (t))

f (x) = x^{5}

\int_{- 1}^{1} 4 x^{3} d x

\int 1 + ln (x) + \frac{y}{x} d x

\int - 0.3 y d y

What is inverse oflaplace 5/(2(s-1)^2+8) ?
The answer to inverse oflaplace 5/(2(s-1)^2+8) is e^t 5/4 sin(2t)