What is (-2A-2Bt+B)e^{-2t} ?

The solution to (-2A-2Bt+B)e^{-2t} is (-2Bt+B-2A)/(e^{2t)}

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simplify (-2A-2Bt+B)e^{-2t}

simplify

(- 2 A - 2 Bt + B) e^{- 2 t}

\frac{- 2 Bt + B - 2 A}{e ^{2 t}}

Solution steps

(- 2 A - 2 Bt + B) e^{- 2 t}

Apply exponent rule:

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

= \frac{1}{e ^{2 t}} (- 2 Bt + B - 2 A)

Multiply fractions:

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

= \frac{1 \cdot ( - 2 A - 2 Bt + B )}{e ^{2 t}}

1 \cdot (- 2 A - 2 Bt + B) = - 2 A - 2 Bt + B

= \frac{- 2 A - 2 Bt + B}{e ^{2 t}}

(42) 32 - (42)^{2}

e^{- 2 l n (c o s (2 x))}

e^{(- 1.0074 \cdot 0.25)}

\frac{( n ^{5} )}{( n + 1 ) ^{5}}

10 \cdot 0.41 \cdot 1 0^{- 5} \cdot 0.46

What is (-2A-2Bt+B)e^{-2t} ?
The solution to (-2A-2Bt+B)e^{-2t} is (-2Bt+B-2A)/(e^{2t)}