2022 H2 Mathematics Paper 1 Question 12 Differential Equations (DEs)
Answers d P d t = k P . {\frac{\mathrm{d}P}{\mathrm{d}t} = kP.} d t d P = k P . P = 50 e t ln 2 10 . {P = 50 \mathop{\mathrm{e}^{\frac{t \ln 2}{10}}}.} P = 50 e 10 t l n 2 . P = 500 9 e − t ln 9 4 10 + 1 . {\displaystyle P = \frac{500}{9\mathrm{e}^{\frac{-t \ln \frac{9}{4}}{10}} + 1}.} P = 9 e 10 − t l n 4 9 + 1 500 . The population approaches
500 {500} 500 in the long run.
This model is an improvement as the first model predicts that the population of the
species approaches
∞ {\infty} ∞ in the long run, which is not realistic. In the long
run, we expect the population of the species to stabilise at a certain value, which is
what this refined model predicts.
Full solutions
(a) d P d t = k P ■ 1 P d P d t = k ∫ 1 P d P = ∫ k d t ln ∣ P ∣ = k t + c ∣ P ∣ = e k t + c P = A e k t \begin{align*}
\frac{\mathrm{d}P}{\mathrm{d}t} &= kP \; \blacksquare \\
\frac{1}{P} \frac{\mathrm{d}P}{\mathrm{d}t} &= k \\
\int \frac{1}{P} \; \mathrm{d}P &= \int k \; \mathrm{d}t
\\ \ln |P| &= kt + c
\\ |P| &= \mathrm{e}^{kt + c}
\\ P &= A\mathrm{e}^{kt}
\end{align*} d t d P P 1 d t d P ∫ P 1 d P ln ∣ P ∣ ∣ P ∣ P = k P ■ = k = ∫ k d t = k t + c = e k t + c = A e k t
When
t = 0 , {t=0,} t = 0 , P = 50 {P=50} P = 50 50 = A e k ( 0 ) A = 50 \begin{align*}
50 &= A \mathrm{e}^{k(0)}
\\ A &= 50
\end{align*} 50 A = A e k ( 0 ) = 50
When
t = 10 , {t=10,} t = 10 , P = 100 {P=100} P = 100 100 = 50 e k ( 10 ) e 10 k = 2 10 k = ln 2 k = 1 10 ln 2 \begin{align*}
100 &= 50 \mathrm{e}^{k(10)}
\\ \mathrm{e}^{10k} &= 2
\\ 10k &= \ln 2
\\ k &= \frac{1}{10} \ln 2
\end{align*} 100 e 10 k 10 k k = 50 e k ( 10 ) = 2 = ln 2 = 10 1 ln 2 P = 50 e t ln 2 10 ■ P = 50 \mathop{\mathrm{e}^{\frac{t \ln 2}{10}}} \; \blacksquare P = 50 e 10 t l n 2 ■
(b) 1 P ( 500 − P ) d P d t = λ ∫ 1 P ( 500 − P ) d P = ∫ λ d t 1 500 ∫ 1 P + 1 500 − P d P = λ t + c ln ∣ P ∣ − ln ∣ 500 − P ∣ = 500 λ t + c ′ ln ∣ P 500 − P ∣ = 500 λ t + c ′ P 500 − P = B e 500 λ t \begin{align*}
\frac{1}{P(500-P)} \frac{\mathrm{d}P}{\mathrm{d}t} &= \lambda
\\ \int \frac{1}{P(500-P)} \; \mathrm{d}P &= \int \lambda \mathrm{d}t
\\ \frac{1}{500} \int \frac{1}{P} + \frac{1}{500-P} \; \mathrm{d}P &= \lambda t + c
\\ \ln |P| - \ln |500-P| &= 500 \lambda t + c'
\\ \ln \left| \frac{P}{500-P} \right| &= 500 \lambda t + c'
\\ \frac{P}{500-P} &= B \mathrm{e}^{500 \lambda t}
\end{align*} P ( 500 − P ) 1 d t d P ∫ P ( 500 − P ) 1 d P 500 1 ∫ P 1 + 500 − P 1 d P ln ∣ P ∣ − ln ∣500 − P ∣ ln 500 − P P 500 − P P = λ = ∫ λ d t = λ t + c = 500 λ t + c ′ = 500 λ t + c ′ = B e 500 λ t
When
t = 0 , {t=0,} t = 0 , P = 50 {P=50} P = 50 50 500 − 50 = B e 500 λ ( 0 ) B = 1 9 \begin{align*}
\frac{50}{500-50} &= B \mathrm{e}^{500 \lambda(0)}
\\ B &= \frac{1}{9}
\end{align*} 500 − 50 50 B = B e 500 λ ( 0 ) = 9 1
When
t = 10 , {t=10,} t = 10 , P = 100 {P=100} P = 100 100 500 − 100 = 1 9 e 500 λ ( 10 ) e 5000 λ = 9 4 5000 λ = ln 9 4 500 λ = 1 10 ln 9 4 \begin{align*}
\frac{100}{500-100} &= \frac{1}{9} \mathrm{e}^{500 \lambda (10)}
\\ \mathrm{e}^{5000 \lambda} &= \frac{9}{4}
\\ 5000 \lambda &= \ln \frac{9}{4}
\\ 500 \lambda &= \frac{1}{10} \ln \frac{9}{4}
\end{align*} 500 − 100 100 e 5000 λ 5000 λ 500 λ = 9 1 e 500 λ ( 10 ) = 4 9 = ln 4 9 = 10 1 ln 4 9 P 500 − P = 1 9 e t ln 9 4 10 9 P = 500 e t ln 9 4 10 − P e t ln 9 4 10 P ( 9 + e t ln 9 4 10 ) = 500 e t ln 9 4 10 \begin{gather*}
\frac{P}{500-P} = \frac{1}{9} \mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}
\\ 9P = 500 \mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}} - P \mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}
\\ P(9 + \mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}) = 500\mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}
\end{gather*} 500 − P P = 9 1 e 10 t l n 4 9 9 P = 500 e 10 t l n 4 9 − P e 10 t l n 4 9 P ( 9 + e 10 t l n 4 9 ) = 500 e 10 t l n 4 9 P = 500 e t ln 9 4 10 9 + e t ln 9 4 10 = 500 9 e − t ln 9 4 10 + 1 ■ \begin{align*}
P &= \frac{500\mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}}{9+\mathrm{e}^{\frac{t \ln \frac{9}{4}}{10}}}
\\ &= \frac{500}{9\mathrm{e}^{\frac{-t \ln \frac{9}{4}}{10}} + 1} \; \blacksquare
\end{align*} P = 9 + e 10 t l n 4 9 500 e 10 t l n 4 9 = 9 e 10 − t l n 4 9 + 1 500 ■
(c) As
t → ∞ , {t \to \infty,} t → ∞ , e − t ln 9 4 10 → 0 {\mathrm{e}^{\frac{-t \ln \frac{9}{4}}{10}} \to 0} e 10 − t l n 4 9 → 0 so
P → 500. {P \to 500.} P → 500.
Hence the population approaches 500 {500} 500 ■ {\blacksquare} ■
This model is an improvement as the first model predicts that the population of the
species approaches ∞ {\infty} ∞ ■ {\blacksquare} ■
Question Commentary
This DE question is very similar to recent years (the last three, in fact!) so hopefully
most students will be able to apply the techniques from their revision comfortably.
The integral for part (b) involves using either partial fractions (the technique shown in the
solution above) or completing the square.