What is inverse oflaplace (s+1)/(s^2+2s+10) ?

The answer to inverse oflaplace (s+1)/(s^2+2s+10) is e^{-t}cos(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (s+1)/(s^2+2s+10)

inverse laplace

\frac{s + 1}{s ^{2} + 2 s + 10}

e^{- t} cos (3 t)

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} + 2 s + 10}}

\frac{s + 1}{s ^{2} + 2 s + 10} = \frac{s + 1}{( s + 1 ) ^{2} + 9}

= L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 9}}

Apply inverse transform rule:

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

then

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

For

\frac{s + 1}{( s + 1 ) ^{2} + 9} : a = - 1, F (s) = \frac{s}{s ^{2} + 9}

= e^{- t} L^{- 1} {\frac{s}{s ^{2} + 9}}

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

= e^{- t} cos (3 t)

What is inverse oflaplace (s+1)/(s^2+2s+10) ?
The answer to inverse oflaplace (s+1)/(s^2+2s+10) is e^{-t}cos(3t)