What is inverse oflaplace (-3)/((s-3)^2+4) ?

The answer to inverse oflaplace (-3)/((s-3)^2+4) is -3/2 e^{3t}sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (-3)/((s-3)^2+4)

inverse laplace

\frac{- 3}{( s - 3 ) ^{2} + 4}

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

Solution steps

L^{- 1} {\frac{- 3}{( s - 3 ) ^{2} + 4}}

Apply inverse transform rule:

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

then

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

For

\frac{- 3}{( s - 3 ) ^{2} + 4} : a = 3, F (s) = \frac{- 3}{s ^{2} + 4}

= e^{3 t} L^{- 1} {\frac{- 3}{s ^{2} + 4}}

L^{- 1} {\frac{- 3}{s ^{2} + 4}} : - \frac{3}{2} sin (2 t)

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

What is inverse oflaplace (-3)/((s-3)^2+4) ?
The answer to inverse oflaplace (-3)/((s-3)^2+4) is -3/2 e^{3t}sin(2t)