What is inverse oflaplace 3/((s^2+1)(s^2+9)) ?

The answer to inverse oflaplace 3/((s^2+1)(s^2+9)) is 3/8 sin(t)-1/8 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 3/((s^2+1)(s^2+9))

inverse laplace

\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )}

\frac{3}{8} sin (t) - \frac{1}{8} sin (3 t)

Solution steps

L^{- 1} {\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )} : \frac{3}{8 ( s ^{2} + 1 )} - \frac{3}{8 ( s ^{2} + 9 )}

= L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )} - \frac{3}{8 ( s ^{2} + 9 )}}

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{3}{8 ( s ^{2} + 1 )}} - L^{- 1} {\frac{3}{8 ( s ^{2} + 9 )}}

L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )}} : \frac{3}{8} sin (t)

L^{- 1} {\frac{3}{8 ( s ^{2} + 9 )}} : \frac{1}{8} sin (3 t)

= \frac{3}{8} sin (t) - \frac{1}{8} sin (3 t)

\int \frac{3 x ^{2} + 2 x + 1}{x ^{2} ( x + 1 )} d x

\int - sin^{2} (x) + cos^{2} (x) d x

\int_{1}^{3} \frac{x ^{2} - 1}{x} d x

\int arctan (x - 1) d x

\int \frac{1}{( x ^{3} - 1 )} d x

What is inverse oflaplace 3/((s^2+1)(s^2+9)) ?
The answer to inverse oflaplace 3/((s^2+1)(s^2+9)) is 3/8 sin(t)-1/8 sin(3t)