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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

e^{- t} \cdot 3 sin (t)

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

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

= e^{- t} \cdot 3 sin (t)

\frac{\partial}{\partial x} (\frac{y}{x} - 1)

f (x) = cos (\frac{x ^{2} - 25}{x - 5})

x \to 0 lim (\frac{2 x + x ^{2}}{x})

\int 4 x^{3} n xx d x

\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}

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