What is inverse oflaplace 1/(s^2-6s+18) ?

The answer to inverse oflaplace 1/(s^2-6s+18) is e^{3t} 1/3 sin(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/(s^2-6s+18)

inverse laplace

\frac{1}{s ^{2} - 6 s + 18}

e^{3 t} \frac{1}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} - 6 s + 18}}

\frac{1}{s ^{2} - 6 s + 18} = \frac{1}{( s - 3 ) ^{2} + 9}

= L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 9}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{1}{( s - 3 ) ^{2} + 9} : a = 3, F (s) = \frac{1}{s ^{2} + 9}

= e^{3 t} L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= e^{3 t} \frac{1}{3} sin (3 t)

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4 y ^{2}})

\frac{d y}{d x} = 4 x + y

y^{^{'}} = y (7 - y) - \frac{49}{4}

\frac{d}{d x} (3 x^{3} - 5 x)

\frac{2}{x ^{2}} + \frac{3}{x ^{3}}

What is inverse oflaplace 1/(s^2-6s+18) ?
The answer to inverse oflaplace 1/(s^2-6s+18) is e^{3t} 1/3 sin(3t)