Page 36 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Solution: (a) Substitute n = 20 and dn = 0.9
dt
1
0.9 = (20)(k – 20)
20k
0.9k = k – 20 ⇒ k = 200
Hence, the maximum growth rate is 2.5 bears per year.Reserved.
1
(b) \ The differential equation is dn = 4000 n(200 – n).
dt
1
The expression 4000 n(200 – n) is a quadratic expression and has the maximum
1
1
(0 + 200) = 100. The maximum value of
n(200 – n) is
value at n =
2
4000
1
4000 (100)(200 – 100) = 2.5

(c) As n approaches 200, the growth rate of the bears approaches 0, i.e. there will
be no bear born.
(d) Separating the variables and integrating both sides:

dn
1
∫ 1 n(200 – n) 2 ∫ 4000 dt

=
1 1 + 1 2 dn = 1 ∫ dt
∫ 1
200 n (200 – n) 4000
n
(ln – ln (200 – n) = 0.05t + C
n
ln 1 200 – n 2 = 0.05t + C
4 Penerbitan Pelangi Sdn Bhd. All Rights
Substituting t = 0, n = 50

50
ln 1 200 – 50 2 = 0.05(0) + C

⇒ C = ln 1
3
n
\ ln 1 200 – n 2 = 0.05t + ln 1
3
n
1
ln 1 200 – n 2 – ln = 0.05t
3
3n
ln 1 200 – n 2 = 0.05t
3n
1 200 – n 2 = e 0.05t
1 200 – n 2 = –0.05t
e
3n
200 – n = 3ne –0.05t

\ n = 200
1 + 3e –0.05t

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04 STPM Math(T) T2.indd 144 28/01/2022 5:44 PM