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

The answer to laplacetransform e^{-t}sin(2t) is 2/((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}sin(2t)

laplace transform

e^{- t} sin (2 t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

\int_{1}^{2} (\frac{1}{x}) d x

\int \frac{sin ( 2 x )}{11 + cos ^{2} ( x )} d x

\int tan^{3} (x) \cdot sec (x) d x

\frac{\partial}{\partial x} (cos (x^{2} \cdot y))

h \to 0 lim (\frac{4 ^{h} - 1}{h})

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