2020 H2 Mathematics Paper 1 Question 10

Differential Equations (DEs)

Answers

(i)

dPdt=−0.03P{\frac{\mathrm{d}P}{\mathrm{d}t} = -0.03 P}

(ii)

P=Ae−0.03t{P = A\mathrm{e}^{-0.03t}}
The number of sheep will approach 0{0} if this situation continues over many years

(iii)

dPdt=n−0.03P{\frac{\mathrm{d}P}{\mathrm{d}t} = n - 0.03P}

(iv)

P=1003(n−Ae−0.03t){P = \frac{100}{3} \left( n-A\mathrm{e}^{-0.03t} \right)}

(v)

n=15{n=15}

Full solutions

(i)

dPdt=−0.03P  ■\frac{\mathrm{d}P}{\mathrm{d}t} = -0.03 P \; \blacksquare

(ii)

1PdPdt=−0.03∫1P  dP=∫−0.03  dtln⁡P=−0.03t+cP=e−0.03t+cP=Ae−0.03t  ■\begin{gather*} \frac{1}{P} \frac{\mathrm{d}P}{\mathrm{d}t} = -0.03 \\ \int \frac{1}{P} \; \mathrm{d}P = \int -0.03 \; \mathrm{d}t \\ \ln P = -0.03t + c \\ P = \mathrm{e}^{-0.03t+c} \\ P = A\mathrm{e}^{-0.03t} \; \blacksquare \end{gather*}
As t→∞,e−0.03t→0{t \to \infty, \mathrm{e}^{-0.03t} \to 0}
P→0P \to 0
Hence the number of sheep will approach 0{0} if this situation continues over many years

(iii)

dPdt=n−0.03P  ■\frac{\mathrm{d}P}{\mathrm{d}t} = n - 0.03P \; \blacksquare

(iv)

∫1n−0.03P  dP=∫1  dt1−0.03ln⁡∣n−0.03P∣=t+C∣n−0.03P∣=e−0.03t−0.03C∣n−0.03P∣=e−0.03Ce−0.03tn−0.03P=Ae−0.03tP=n−Ae−0.03t0.03P=1003(n−Ae−0.03t)  ■\begin{gather*} \int \frac{1}{n-0.03P} \; \mathrm{d}P = \int 1 \; \mathrm{d}t \\ \frac{1}{-0.03} \ln |n-0.03P| = t + C \\ |n-0.03P| = \mathrm{e}^{-0.03t-0.03C} \\ |n-0.03P| = \mathrm{e}^{-0.03C}\mathrm{e}^{-0.03t} \\ n-0.03P = A\mathrm{e}^{-0.03t} \\ P = \frac{n-A\mathrm{e}^{-0.03t}}{0.03} \\ P = \frac{100}{3} \left( n-A\mathrm{e}^{-0.03t} \right) \; \blacksquare \end{gather*}

(v)

As t→∞,e−0.03t→0{t \to \infty, \mathrm{e}^{-0.03t} \to 0}
P→1003n500=1003nn=15  ■\begin{align*} P &\to \frac{100}{3}n \\ 500 &= \frac{100}{3}n \\ n &= 15 \; \blacksquare \end{align*}