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} ( 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)

e^{2 t} cos (3 t - 5)

f (x) = 2 x^{4} - 11 x^{2}, at x = 2

f (x) = \frac{x ^{2} + 3 x + 2}{x ^{2} - 1}

x^{2} y^{^{''}} - 6 x y^{^{'}} + 10 y = - 8 x^{3}

\int_{- 4}^{4} \frac{1}{2} (1 6^{2} - x^{4}) d 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)