What is inverse oflaplace 1/(2002s^2+9) ?

The answer to inverse oflaplace 1/(2002s^2+9) is 1/(3sqrt(2002))sin((3t)/(sqrt(2002)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(2002s^2+9)

inverse laplace

\frac{1}{2002 s ^{2} + 9}

\frac{1}{3 2002} sin (\frac{3 t}{2002})

Solution steps

L^{- 1} {\frac{1}{2002 s ^{2} + 9}}

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{3 2002} \cdot \frac{\frac{3}{2002}}{s ^{2} + ( \frac{3}{2002} ) ^{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}{3 2002} L^{- 1} ⎩ ⎨ ⎧ \frac{\frac{3}{2002}}{s ^{2} + ( \frac{3}{2002} ) ^{2}} ⎭ ⎬ ⎫

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{\frac{3}{2002}}{s ^{2} + ( \frac{3}{2002} ) ^{2}}} = sin (\frac{3}{2002} t)

= \frac{1}{3 2002} sin (\frac{3}{2002} t)

= \frac{1}{3 2002} sin (\frac{3 t}{2002})

y^{^{'}} = \frac{x ^{3} y ^{3} + 4 x ^{3}}{y ^{2}}

\frac{d}{d x} (x^{2} + y^{2}) = 25

\frac{\partial}{\partial x} (9 x^{2})

f (x) = 4 - x^{2}

x \to - 2 lim (\frac{x ^{2} - 1}{2 - x})

What is inverse oflaplace 1/(2002s^2+9) ?
The answer to inverse oflaplace 1/(2002s^2+9) is 1/(3sqrt(2002))sin((3t)/(sqrt(2002)))