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

The answer to inverse oflaplace 1/((s+1)^2+5) is e^{-t} 1/(sqrt(5))sin(sqrt(5)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

e^{- t} \frac{1}{5} sin (5 t)

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

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

= e^{- t} \frac{1}{5} sin (5 t)

\frac{h ^{2} - 2 h + 5}{8}

3 x^{\frac{1}{4}} (x^{3} - 5)

\int_{0}^{\infty} sin (5 x) d x

\frac{\partial}{\partial x} (e^{2 x} y)

\frac{\partial}{\partial x} (x^{2} + 4)

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