What is inverse oflaplace 1/((s-3)^2+3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/((s-3)^2+3)

inverse laplace

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

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

Solution steps

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

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

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

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

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

\int \frac{35 x ^{4} - 3}{7 x ^{5} - 3 x} d x

(- 7, - 3), (- 5, - 8)

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

\frac{d}{d x} ((x - 2) (x + 1) (x + 5))

\int (x^{2} - 4 x - 15) d x

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