What is laplacetransform 5e^{-3t}sin(t) ?

The answer to laplacetransform 5e^{-3t}sin(t) is 5/((s+3)^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform 5e^{-3t}sin(t)

laplace transform

5 e^{- 3 t} sin (t)

\frac{5}{( s + 3 ) ^{2} + 1}

Solution steps

L {5 e^{- 3 t} sin (t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 5 L {e^{- 3 t} sin (t)}

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

= 5 \cdot \frac{1}{( s + 3 ) ^{2} + 1}

Refine

5 \frac{1}{( s + 3 ) ^{2} + 1} : \frac{5}{( s + 3 ) ^{2} + 1}

= \frac{5}{( s + 3 ) ^{2} + 1}

\int cos (16 x) cos (7 x) d x

\frac{d ^{2}}{d x ^{2}} (x^{r})

\frac{1}{s ^{2} + 5}

n = 0 \sum \infty (n + 1) (\frac{1}{2})^{n}

\frac{1}{1 - x ^{5}}

What is laplacetransform 5e^{-3t}sin(t) ?
The answer to laplacetransform 5e^{-3t}sin(t) is 5/((s+3)^2+1)