What is inverse oflaplace 8/(s^2+9) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 8/(s^2+9)

inverse laplace

\frac{8}{s ^{2} + 9}

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

Solution steps

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

= L^{- 1} {\frac{8}{3} \cdot \frac{3}{s ^{2} + 3 ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

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

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

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

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