What is inverse oflaplace (5s)/((s^2+4)(s+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (5s)/((s^2+4)(s+1))

inverse laplace

\frac{5 s}{( s ^{2} + 4 ) ( s + 1 )}

cos (2 t) + 2 sin (2 t) - e^{- t}

Solution steps

L^{- 1} {\frac{5 s}{( s ^{2} + 4 ) ( s + 1 )}}

Take the partial fraction of

\frac{5 s}{( s ^{2} + 4 ) ( s + 1 )} : \frac{s + 4}{s ^{2} + 4} - \frac{1}{s + 1}

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

= cos (2 t) + 4 \cdot \frac{1}{2} sin (2 t) - e^{- t}

Refine

cos (2 t) + 4 \frac{1}{2} sin (2 t) - e^{- t} : cos (2 t) + 2 sin (2 t) - e^{- t}

= cos (2 t) + 2 sin (2 t) - e^{- t}

\frac{d y}{d x} = 0.002 y (500 - y) - 80

\frac{d}{d x} (x - 16)

\int e^{- 0.1 t} d t

\int_{0}^{\infty} e^{2 x} d x

\frac{d}{d t} (5 t)

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