Core Pure · Modelling with Differential Equations · Year 2

First-Order Modelling

Translating rates of change — mixing problems, growth models and kinematics — into solvable differential equations.

14 min read 3 worked examples

Building a differential equation from a real situation

Method — translating "rate of change" statements

Phrases like "the rate of change of $x$ is proportional to..." translate directly into $\dfrac{dx}{dt}=k\times(\ldots)$. Where a quantity is being added and removed simultaneously (a mixing or flow problem), the net rate is (rate in) $-$ (rate out); each of these may itself depend on the current amount present.

Example 1 · A mixing (tank) problem

A tank contains 800 litres of pure water. Salt solution, with concentration $0.2\text{ kg}$ of salt per litre, is added at a constant rate of $5$ litres per minute, and the well-mixed solution drains from the tank at the same rate. Given that there are $x$ kg of salt in the tank after $t$ minutes, show that $\dfrac{dx}{dt}=1-\dfrac{x}{160}$, and find $x$ in terms of $t$.

Setting up: salt enters at rate $5\times0.2=1$ kg/min. Since the volume stays constant at $800$ litres (inflow and outflow rates match), the concentration in the tank at time $t$ is $\dfrac x{800}$ kg/litre, so salt leaves at rate $5\times\dfrac x{800}=\dfrac x{160}$ kg/min.

Net rate: $\dfrac{dx}{dt}=1-\dfrac x{160}$, as required.

Solving (linear, integrating factor $I=e^{t/160}$): $\dfrac{d}{dt}\left(xe^{t/160}\right)=e^{t/160}$.

Integrate: $xe^{t/160}=160e^{t/160}+c$.

At $t=0$, $x=0$ (pure water initially): $0=160+c\Rightarrow c=-160$.

So $x=160-160e^{-t/160}=160\left(1-e^{-t/160}\right)$.

Kinematics with v dv/dx

Method — acceleration as a function of displacement

When acceleration is given as a function of displacement $x$ rather than time, use $\dfrac{d^2x}{dt^2}=v\dfrac{dv}{dx}$ (from the chain rule, since $v=\dfrac{dx}{dt}$) to turn the equation into one relating $v$ and $x$ only, which can then often be solved by separating variables.

ProofWhy d²x/dt² = v dv/dx

By the chain rule, $\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}$.

Since $\dfrac{dx}{dt}=v$, this gives $\dfrac{dv}{dt}=v\dfrac{dv}{dx}$.

And $\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}$ by definition of acceleration, so $\dfrac{d^2x}{dt^2}=v\dfrac{dv}{dx}$.

∎

Example 2 · Using v dv/dx

A particle moves in a straight line such that $\dfrac{d^2x}{dt^2}=-9x$. Given that $v=0$ when $x=5$, find $v^2$ in terms of $x$.

Rewrite using $\dfrac{d^2x}{dt^2}=v\dfrac{dv}{dx}$: $v\dfrac{dv}{dx}=-9x$.

Separate variables: $\displaystyle\int v\,dv=\int-9x\,dx$, so $\dfrac{v^2}2=-\dfrac{9x^2}2+c$.

At $x=5$, $v=0$: $0=-\dfrac{9(25)}2+c\Rightarrow c=\dfrac{225}2$.

So $v^2=225-9x^2$.

Other rate-of-change models

Example 3 · A growth-rate model

The population $P$ of bacteria in a culture grows at a rate proportional to $P(1000-P)$, where $1000$ is the maximum sustainable population. Given that $P=100$ when $t=0$ and $P=500$ when $t=2$, form and solve the differential equation, giving $P$ in terms of $t$.

$\dfrac{dP}{dt}=kP(1000-P)$.

Separate: $\displaystyle\int\frac1{P(1000-P)}\,dP=\int k\,dt$. Using partial fractions, $\dfrac1{P(1000-P)}=\dfrac1{1000}\left(\dfrac1P+\dfrac1{1000-P}\right)$.

$$\frac1{1000}\Big(\ln P-\ln(1000-P)\Big)=kt+c \;\Rightarrow\; \ln\left(\frac P{1000-P}\right)=1000kt+c_1$$

At $t=0$, $P=100$: $\ln\left(\dfrac{100}{900}\right)=c_1\Rightarrow c_1=\ln\dfrac19$.

At $t=2$, $P=500$: $\ln\left(\dfrac{500}{500}\right)=2000k+\ln\dfrac19\Rightarrow0=2000k+\ln\dfrac19\Rightarrow k=\dfrac{\ln9}{2000}$.

So $\ln\left(\dfrac P{1000-P}\right)=\dfrac{t\ln9}2+\ln\dfrac19$, which rearranges (taking exponentials) to $\dfrac P{1000-P}=\dfrac19\times9^{t/2}=9^{t/2-1}$, giving $P=\dfrac{1000\times9^{t/2-1}}{1+9^{t/2-1}}$.

Exam-style questions

Question 1 7 marks

A tank contains $500$ litres of water with $20$ kg of dissolved chemical. Pure water flows in at a rate of $10$ litres per minute, and the well-mixed solution flows out at the same rate. Given that there are $x$ kg of chemical after $t$ minutes, form a differential equation for $x$, and solve it to find $x$ in terms of $t$.

No chemical enters (pure water); it leaves at rate $10\times\dfrac x{500}=\dfrac x{50}$, so $\dfrac{dx}{dt}=-\dfrac x{50}$. (M1, A1)

Separate: $\displaystyle\int\frac1x\,dx=\int-\frac1{50}\,dt \Rightarrow \ln x=-\frac t{50}+c$. (M1, A1)

At $t=0,x=20$: $c=\ln20$, so $x=20e^{-t/50}$. (M1, A1, A1)

Question 2 6 marks

A particle moves in a straight line such that $\dfrac{d^2x}{dt^2}=\dfrac{2}{x^2}$. Given that $v=3$ when $x=1$, find $v^2$ in terms of $x$.

$v\dfrac{dv}{dx}=\dfrac2{x^2}$. (M1)

$\displaystyle\int v\,dv=\int2x^{-2}\,dx \Rightarrow \frac{v^2}2=-\frac2x+c$. (M1, A1)

At $x=1,v=3$: $\dfrac92=-2+c\Rightarrow c=\dfrac{13}2$. (M1, A1)

$v^2=13-\dfrac4x$. (A1)

Question 3 8 marks

A pond initially contains $2000$ fish. The population $N$ is modelled by $\dfrac{dN}{dt}=0.001N(4000-N)$. Given that $N=2000$ when $t=0$, show that $\dfrac N{4000-N}=e^{4t}$, and hence find $N$ in terms of $t$.

$\displaystyle\int\frac1{N(4000-N)}\,dN=\int0.001\,dt$; partial fractions give $\dfrac1{4000}\left(\dfrac1N+\dfrac1{4000-N}\right)$. (M1, A1)

$\dfrac1{4000}\big(\ln N-\ln(4000-N)\big)=0.001t+c \Rightarrow \ln\left(\dfrac N{4000-N}\right)=4t+c_1$. (M1, A1)

At $t=0,N=2000$: $\ln\left(\dfrac{2000}{2000}\right)=\ln1=0=c_1$. (M1)

So $\dfrac N{4000-N}=e^{4t}$, as required. (A1)

Rearranging: $N=e^{4t}(4000-N)\Rightarrow N(1+e^{4t})=4000e^{4t}\Rightarrow N=\dfrac{4000e^{4t}}{1+e^{4t}}$. (A1)

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Differential Equations: Boundary Conditions

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