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

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

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

simplify

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

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

Solution steps

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

Apply exponent rule:

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

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

Multiply fractions:

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

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

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

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

(1 + sin (2 π x)) e^{- \frac{x}{10}}

(3 a^{3} b^{3}) (16 a^{2} b^{4})

(2 x^{2} y^{3}) (3 x z^{2}) (5 y)

e^{- 0.46209 \cdot 16}

(1 - x^{2})^{x}

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