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

The answer to inverse oflaplace {1/((s^2+1)(s^2+4))} is 1/3 sin(t)-1/6 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+1)(s^2+4))}

inverse laplace

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

\frac{1}{3} sin (t) - \frac{1}{6} sin (2 t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 4 )} : \frac{1}{3 ( s ^{2} + 1 )} - \frac{1}{3 ( s ^{2} + 4 )}

= L^{- 1} {\frac{1}{3 ( s ^{2} + 1 )} - \frac{1}{3 ( 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}{3 ( s ^{2} + 1 )}} - L^{- 1} {\frac{1}{3 ( s ^{2} + 4 )}}

L^{- 1} {\frac{1}{3 ( s ^{2} + 1 )}} : \frac{1}{3} sin (t)

L^{- 1} {\frac{1}{3 ( s ^{2} + 4 )}} : \frac{1}{6} sin (2 t)

= \frac{1}{3} sin (t) - \frac{1}{6} sin (2 t)

x \to 0 lim (\frac{sin ( 4 x )}{2})

\int x^{2} - 10 d x

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

\frac{d}{d x} (2 e^{2 x} x)

\int e^{3 x + 1} d x

What is inverse oflaplace {1/((s^2+1)(s^2+4))} ?
The answer to inverse oflaplace {1/((s^2+1)(s^2+4))} is 1/3 sin(t)-1/6 sin(2t)