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

The answer to inverse oflaplace 1/(s(s^2+4))+1 is 1/4 \H(t)-1/4 cos(2t)+\delta(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(s^2+4))+1

inverse laplace

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

\frac{1}{4} H (t) - \frac{1}{4} cos (2 t) + δ (t)

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

L^{- 1} {1} : δ (t)

= \frac{1}{4} H (t) - \frac{1}{4} cos (2 t) + δ (t)

x \to 0 lim (\frac{\frac{4}{x - 1} + 4}{x})

\frac{d}{d x} (\frac{x}{x + 2})

e^{4 x s i n (2 x)}

f (x) = x^{3} + 5 x

\frac{\partial}{\partial x} (4 e^{4 x y})

What is inverse oflaplace 1/(s(s^2+4))+1 ?
The answer to inverse oflaplace 1/(s(s^2+4))+1 is 1/4 \H(t)-1/4 cos(2t)+\delta(t)