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

The answer to inverse oflaplace 2/((s-1)^2) is e^t*2t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 2/((s-1)^2)

inverse laplace

\frac{2}{( s - 1 ) ^{2}}

e^{t} \cdot 2 t

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

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

= e^{t} \cdot 2 t

\int \frac{9 x}{3 x + 1} d x

f (x) = \frac{x}{x + 3}, at x = 4

\frac{1}{s ( s - 2 ) ^{2}}

y = \frac{x}{64 + x ^{2}}, (0, 0)

\frac{d}{d x} (2 sin (x) - 1)

What is inverse oflaplace 2/((s-1)^2) ?
The answer to inverse oflaplace 2/((s-1)^2) is e^t*2t