What is inverse oflaplace 1/(s^2+4) ?

The answer to inverse oflaplace 1/(s^2+4) is 1/2 sin(2t)

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+4)

inverse laplace

\frac{1}{s ^{2} + 4}

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

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 4}}

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

Use Inverse Laplace Transform table:

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

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

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

7 t \cdot \frac{d y}{d t} - 4 y = t

\frac{d}{d x} (1 - sin (3 x))

\frac{e ^{- 5 s}}{s ^{2}}

x \to 3 + lim (\frac{( x ^{2} )}{x - 3})

What is inverse oflaplace 1/(s^2+4) ?
The answer to inverse oflaplace 1/(s^2+4) is 1/2 sin(2t)