What is laplacetransform e^{-t}cos(4t)+t/5 ?

The answer to laplacetransform e^{-t}cos(4t)+t/5 is (s+1)/((s+1)^2+16)+1/(5s^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform e^{-t}cos(4t)+t/5

laplace transform

e^{- t} cos (4 t) + \frac{t}{5}

\frac{s + 1}{( s + 1 ) ^{2} + 16} + \frac{1}{5 s ^{2}}

Solution steps

L {e^{- t} cos (4 t) + \frac{t}{5}}

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^{- t} cos (4 t)} + \frac{1}{5} L {t}

L {e^{- t} cos (4 t)} : \frac{s + 1}{( s + 1 ) ^{2} + 16}

L {t} : \frac{1}{s ^{2}}

= \frac{s + 1}{( s + 1 ) ^{2} + 16} + \frac{1}{5} \cdot \frac{1}{s ^{2}}

Refine

\frac{s + 1}{( s + 1 ) ^{2} + 16} + \frac{1}{5} \frac{1}{s ^{2}} : \frac{s + 1}{( s + 1 ) ^{2} + 16} + \frac{1}{5 s ^{2}}

= \frac{s + 1}{( s + 1 ) ^{2} + 16} + \frac{1}{5 s ^{2}}

x \to 2 lim (sin (x - 2))

\frac{\partial}{\partial x} (50 - x^{2} - y^{2})

\frac{d}{d x} (x e^{x} + 1)

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

x \to + (- 1) lim (arccos (x))

What is laplacetransform e^{-t}cos(4t)+t/5 ?
The answer to laplacetransform e^{-t}cos(4t)+t/5 is (s+1)/((s+1)^2+16)+1/(5s^2)