M1 Maths 不定積分24A only http://upload.lsforum.net/users/public/w298581w40.jpg http://upload.lsforum.net/users/public/e202882z40.jpg?

令 u = 100 - p
故 du = - dp

又 p% 表示百分率, 所以 0 ≦ p ≦ 100
因此 u ≧ 0

dp/dt = k( 100 - p )
- du/dt = ku
du / u = - k dt
∫ (1/u) du = ∫ - k dt + c
ln ∣u∣ = ln u = - kt + c , 因為 u ≧ 0 , 所以 ∣u∣ = u
u = e^( - kt + c ) = e^c * e^(-kt) = c' * e^(-kt)
100 - p = c' * e^(-kt) ..... (1)

題目給的初始條件:
廣告尚未推出時, 沒有市民認識該產品, 即
t = 0 時, p = 0
將初始條件代回 (1) 得:
100 - 0 = c' * e^0 = c'
c' = 100

100 - p = 100*e^(-kt)
p = 100 - 100*e^(-kt) = 100*[ 1 - e^(-kt) ] ..... Ans