What is inverse oflaplace s/(s^2+25) ?

The answer to inverse oflaplace s/(s^2+25) is cos(5t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace s/(s^2+25)

inverse laplace

\frac{s}{s ^{2} + 25}

cos (5 t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 25}}

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{s}{s ^{2} + 25} : F (s) = \frac{1}{s ^{2} + 25}

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 25}}) + L^{- 1} {\frac{1}{s ^{2} + 25}} (0)

L^{- 1} {\frac{1}{s ^{2} + 25}} : \frac{1}{5} sin (5 t)

\frac{d}{d t} (\frac{1}{5} sin (5 t)) = cos (5 t)

Evaluate

\frac{1}{5} sin (5 t)

at point

0 = 0

= cos (5 t)

What is inverse oflaplace s/(s^2+25) ?
The answer to inverse oflaplace s/(s^2+25) is cos(5t)