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

The answer to inverse oflaplace 1/(s^2*(s^2+4)) is t/4-1/8 sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

\frac{t}{4} - \frac{1}{8} sin (2 t)

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{4 s ^{2}}} : \frac{t}{4}

L^{- 1} {\frac{1}{4 ( s ^{2} + 4 )}} : \frac{1}{8} sin (2 t)

= \frac{t}{4} - \frac{1}{8} sin (2 t)

\frac{\partial}{\partial x} (sin (5 x - 9 y))

\frac{d}{d x} (x^{5} + 3)

y = x^{2} + x + 1

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

x \to 0 - lim (\frac{7}{1 + e ^{\frac{1}{x}}})

What is inverse oflaplace 1/(s^2*(s^2+4)) ?
The answer to inverse oflaplace 1/(s^2*(s^2+4)) is t/4-1/8 sin(2t)