What is inverse oflaplace 1/((s^2+1)(s^2+3)) ?

The answer to inverse oflaplace 1/((s^2+1)(s^2+3)) is 1/2 sin(t)-1/(2sqrt(3))sin(sqrt(3)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^2+1)(s^2+3))

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

ln ∣ t ∣

\int \frac{4 x ^{3}}{x ^{2} + x} d x

\frac{( x + 1 )}{x - 1}

y = sin (2 x) ln (x^{2}), (1, 0)

\int \frac{v}{1 + v ^{2}} d v

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