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

The solution to e^{-2t(2cos(5t)-sin(5t))} is 1/(e^{2t(2cos(5t)-sin(5t)))}

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

simplify

e^{- 2 t (2 c o s (5 t) - s i n (5 t))}

\frac{1}{e ^{2 t (2 c o s (5 t) - s i n (5 t))}}

Solution steps

e^{- 2 t (2 c o s (5 t) - s i n (5 t))}

Apply exponent rule:

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

= \frac{1}{e ^{2 t (2 c o s (5 t) - s i n (5 t))}}

x 1 + x

e^{- 5 t} sin (2 t)

2 y (x - y)

e^{- l n (n) \cdot t}

(0.5 t^{2} + t) e^{- 2 t}

What is e^{-2t(2cos(5t)-sin(5t))} ?
The solution to e^{-2t(2cos(5t)-sin(5t))} is 1/(e^{2t(2cos(5t)-sin(5t)))}