What is laplacetransform e^{-2t}cos(3t)+5e^{-2t}sin(3t) ?

The answer to laplacetransform e^{-2t}cos(3t)+5e^{-2t}sin(3t) is (s+2)/((s+2)^2+9)+(15)/((s+2)^2+9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform e^{-2t}cos(3t)+5e^{-2t}sin(3t)

laplace transform

e^{- 2 t} cos (3 t) + 5 e^{- 2 t} sin (3 t)

\frac{s + 2}{( s + 2 ) ^{2} + 9} + \frac{15}{( s + 2 ) ^{2} + 9}

Solution steps

L {e^{- 2 t} cos (3 t) + 5 e^{- 2 t} sin (3 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {e^{- 2 t} cos (3 t)} + 5 L {e^{- 2 t} sin (3 t)}

L {e^{- 2 t} cos (3 t)} : \frac{s + 2}{( s + 2 ) ^{2} + 9}

L {e^{- 2 t} sin (3 t)} : \frac{3}{( s + 2 ) ^{2} + 9}

= \frac{s + 2}{( s + 2 ) ^{2} + 9} + 5 \cdot \frac{3}{( s + 2 ) ^{2} + 9}

Refine

\frac{s + 2}{( s + 2 ) ^{2} + 9} + 5 \frac{3}{( s + 2 ) ^{2} + 9} : \frac{s + 2}{( s + 2 ) ^{2} + 9} + \frac{15}{( s + 2 ) ^{2} + 9}

= \frac{s + 2}{( s + 2 ) ^{2} + 9} + \frac{15}{( s + 2 ) ^{2} + 9}

x \to 2 - lim (\frac{x - 2}{x ^{2} - 4})

\frac{x ^{2} - 2}{x ^{2} + 2}

f (x) = 5 3 x

\int sin (6 x) cos (8 x) d x

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

What is laplacetransform e^{-2t}cos(3t)+5e^{-2t}sin(3t) ?
The answer to laplacetransform e^{-2t}cos(3t)+5e^{-2t}sin(3t) is (s+2)/((s+2)^2+9)+(15)/((s+2)^2+9)