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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

2 cos (3 t)

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{2 s}{s ^{2} + 9} : F (s) = \frac{2}{s ^{2} + 9}

= \frac{d}{d t} (L^{- 1} {\frac{2}{s ^{2} + 9}}) + L^{- 1} {\frac{2}{s ^{2} + 9}} (0)

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

\frac{d}{d t} (\frac{2}{3} sin (3 t)) = 2 cos (3 t)

Evaluate

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

at point

0 = 0

= 2 cos (3 t)

y^{^{'}} = 2 - (\frac{x}{20})

y = 2 x, y = \frac{x}{2}

x \to - 6 lim (\frac{x ^{2} + 8}{x + 6})

\frac{d}{d x} (\frac{cos ( ln ( x ) )}{x})

x \to 3 lim (3 x - 3 a x + 2 a)

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