2009 H2 Mathematics Paper 2 Question 4

Differential Equations (DEs)

Answers

(i)

n=5t2−t3+ct+100{n = 5 t^2 - t^3 + ct + 100}

(ii)

n=50(3−e−0.02t){n = 50\left( 3 - \mathrm{e}^{-0.02t} \right)}
The population approaches 150{150 } thousand eventually

Full solutions

(i)

d2ndt2=10−6tdndt=10t−3t2+cn=5t2−t3+ct+d\begin{align*} \frac{\mathrm{d}^{2}n}{\mathrm{d}t^{2}} &= 10 - 6 t \\ \frac{\mathrm{d}n}{\mathrm{d}t} &= 10 t - 3 t^2 + c \\ n &= 5 t^2 - t^3 + ct + d \end{align*}
When t=0,{t=0, } n=100,{n=100,}
100=5(0)2−03+c(0)+dd=100\begin{gather*} 100 = 5(0)^2 - 0^3 + c(0) + d \\ d = 100 \end{gather*}
General solution:
n=5t2−t3+ct+100  ■n = 5 t^2 - t^3 + ct + 100 \; \blacksquare

(ii)

dndt=3−0.02n13−0.02ndndt=1∫13−0.02n  dn=∫1  dt−10.02ln⁡∣3−0.02n∣=t+Cln⁡∣3−0.02n∣=−0.02t−0.02C∣3−0.02n∣=e−0.02t−0.02C∣3−0.02n∣=e−0.02Ce−0.02t3−0.02n=Ae−0.02t\begin{gather*} \frac{\mathrm{d}n}{\mathrm{d}t} = 3-0.02n \\ \frac{1}{3-0.02n} \frac{\mathrm{d}n}{\mathrm{d}t} = 1 \\ \int \frac{1}{3-0.02n} \; \mathrm{d}n = \int 1 \; \mathrm{d}t \\ -\frac{1}{0.02} \ln | 3 - 0.02n | = t + C \\ \ln | 3 - 0.02n | = -0.02t - 0.02C \\ | 3 - 0.02n | = \mathrm{e}^{-0.02t - 0.02C} \\ | 3 - 0.02n | = \mathrm{e}^{- 0.02C} \mathrm{e}^{-0.02t} \\ 3 - 0.02n = A \mathrm{e}^{- 0.02t} \end{gather*}
When t=0,{t=0, } n=100,{n=100,}
3−0.02(100)=Ae−0.02(0)A=1\begin{align*} 3-0.02(100) &= A \mathrm{e}^{-0.02(0)} \\ A &= 1 \end{align*}
3−0.02n=e−0.02tn=50(3−e−0.02t)  ■\begin{gather*} 3 - 0.02n = \mathrm{e}^{- 0.02t} \\ n = 50\left( 3 - \mathrm{e}^{-0.02t} \right) \; \blacksquare \end{gather*}
As t→∞,e−0.02t→0{t \to \infty, \mathrm{e}^{-0.02t} \to 0}
n=50(3−e−0.02t)→150\begin{align*} n &= 50\left( 3 - \mathrm{e}^{-0.02t} \right) \\ &\to 150 \end{align*}
Hence the population approaches 150{150 } thousand eventually ■{\blacksquare}