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

The solution to e^{-2t}(5cos(st)-2sin(3t)+2t) is (2t+5cos(ts)-2sin(3t))/(e^{2t)}

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

simplify

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

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

Solution steps

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

Apply exponent rule:

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

= \frac{1}{e ^{2 t}} (2 t + 5 cos (t s) - 2 sin (3 t))

Multiply fractions:

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

= \frac{1 \cdot ( 5 cos ( s t ) - 2 sin ( 3 t ) + 2 t )}{e ^{2 t}}

1 \cdot (5 cos (s t) - 2 sin (3 t) + 2 t) = 5 cos (t s) - 2 sin (3 t) + 2 t

= \frac{5 cos ( t s ) - 2 sin ( 3 t ) + 2 t}{e ^{2 t}}

263.95 \cdot e^{- 0.027 \cdot 29.2}

(4 a) (3 a^{2})

e^{- \frac{90}{100}}

(e^{2}) (e^{5})

2.51 \cdot 1 0^{- 11}

What is e^{-2t}(5cos(st)-2sin(3t)+2t) ?
The solution to e^{-2t}(5cos(st)-2sin(3t)+2t) is (2t+5cos(ts)-2sin(3t))/(e^{2t)}