What is e^{-at}cos(ωt+θ)u(t) ?

The solution to e^{-at}cos(ωt+θ)u(t) is (tucos(ωt+θ))/(e^{at)}

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simplify e^{-at}cos(ωt+θ)u(t)

simplify

e^{- a t} cos (ω t + θ) u (t)

\frac{t u cos ( ω t + θ )}{e ^{a t}}

Solution steps

e^{- a t} cos (ω t + θ) u (t)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{a t}} t u cos (ω t + θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot cos ( ω t + θ ) u t}{e ^{a t}}

Multiply:

1 \cdot cos (ω t + θ) = cos (ω t + θ)

= \frac{t u cos ( ω t + θ )}{e ^{a t}}

(4 y + 20)^{\circ} (4 y - 80)^{\circ}

\frac{0.02 6 ^{1.85}}{12 0 ^{1.85}}

10 (50 - 100) 0. 1^{- 5}

2 \cdot \frac{2}{2} \cdot e^{- (\frac{2}{2})^{2}}

y^{1}

What is e^{-at}cos(ωt+θ)u(t) ?
The solution to e^{-at}cos(ωt+θ)u(t) is (tucos(ωt+θ))/(e^{at)}