展開する (te^{(-2t)}+sin(t)+cos(t))^2U(t+6)

解

展開する

(t e^{(- 2 t)} + sin (t) + cos (t))^{2} U (t + 6)

e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)

解答ステップ

(t e^{(- 2 t)} + sin (t) + cos (t))^{2} U (t + 6)

= U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} (t + 6)

分配法則を適用する:

a (b + c) = ab + a c

a = (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U, b = t, c = 6

= (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U t + (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U \cdot 6

= t U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} + 6 U (e^{(- 2 t)} t + sin (t) + cos (t))^{2}

簡素化

t U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} + 6 U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} : e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)

= e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)