What is laplacetransform e^{-t}cos(2t) ?

The answer to laplacetransform e^{-t}cos(2t) is (s+1)/((s+1)^2+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform e^{-t}cos(2t)

laplace transform

e^{- t} cos (2 t)

\frac{s + 1}{( s + 1 ) ^{2} + 4}

Solution steps

L {e^{- t} cos (2 t)}

Apply transform rule:

L {f (t)} = F (s)

then

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

For

e^{- t} cos (2 t) : f (t) = cos (2 t), a = - 1

= L {cos (2 t)} (s + 1)

L {cos (2 t)} : \frac{s}{s ^{2} + 4}

= \frac{( s + 1 )}{( s + 1 ) ^{2} + 4}

= \frac{s + 1}{( s + 1 ) ^{2} + 4}

x \to 0 + lim ((\frac{1}{5})^{\frac{1}{x}})

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

\int_{0}^{6} 1 + x d x

x \to 3 + lim (\frac{x + 9}{x - 3})

\int \frac{ln ( x + 8 )}{x + 8} d x

What is laplacetransform e^{-t}cos(2t) ?
The answer to laplacetransform e^{-t}cos(2t) is (s+1)/((s+1)^2+4)