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

The answer to inverse oflaplace 1/(s^2+pi^2) is 1/pi sin(pit)

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+pi^2)

inverse laplace

\frac{1}{s ^{2} + π ^{2}}

\frac{1}{π} sin (π t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + π ^{2}}}

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{π} sin (π t)

x^{2}, - x + 6

(6 + x) y^{^{'}} = 3 y

2 x^{2} - 3 x

f (x) = - 2

\frac{d}{d x} (3 + x)

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