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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 2/(s^2+2)

inverse laplace

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

2 sin (2 t)

Solution steps

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

= L^{- 1} {2 \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)}

= 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)

= 2 sin (2 t)

\frac{d}{d x} (x (x^{4} + 5)^{5})

\int_{0}^{\infty} a e^{- a x} d x

n = 0 \sum \infty ln (n)

n = 0 \sum \infty \frac{3}{1 0 ^{n}}

x^{2} y^{^{''}} - 4 x y^{^{'}} + 4 y = 0

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