G6 - Constructing differential equations | AQA A-Level Mathematics

Translate rate and proportionality language into differential equations, calibrate constants from measurements and state initial conditions. Check signs, units and assumptions, and verify a proposed function by differentiation.

Translating rate and proportionality language

Can an equation describe a curve before we know its formula? Suppose the curve passes through $(0,3)$ and its tangent gradient at $(x,y)$ is $2x+y$ . The statement already tells us how the curve changes: $dy/dx=2x+y$ .

You already know

A derivative is a rate of change: $dy/dx$ is the rate of $y$ with respect to $x$ . A tangent gradient need not involve time. The derivative interpretation is recalled, rather than supplied as a formula.

A differential equation involves a derivative of an unknown function. An initial condition specifies the function's value at a starting input, such as $y(0)=3$ . It is different from $y'(0)=3$ , which specifies an initial gradient.

“Proportional to” introduces a constant multiplier. For example, a mass whose loss rate is proportional to its square root obeys $dm/dt=-k\sqrt m$ , with $k>0$ and $m\ge0$ . The minus sign gives a decreasing mass, while $k\sqrt m$ is the positive magnitude of the loss. This translation is not supplied in the formula booklet.

Translating proportionality

If a signed rate is proportional to $F$ , write

\text{rate}=kF

State the sign of $k$ , or use a positive $k$ and an explicit minus sign for a decrease. Not in the formula booklet: learn it.

Write the changing quantity on top of the derivative and the independent variable below it. Then translate the rate statement before using any numerical data.

Worked example 1: a curve's slope law

Question. A curve passes through $(0,3)$ . Its gradient at $(x,y)$ is $2x+y$ . Write a differential equation and initial condition, then find the initial gradient.

Plan. Replace the word gradient by $dy/dx$ and substitute the starting point.

\begin{aligned} \frac{dy}{dx}&=2x+y && \text{translate the gradient}\\ y(0)&=3 && \text{initial point}\\ y'(0)&=2(0)+3=3 && \text{evaluate the rate} \end{aligned}

Why keep the condition separate? The equation gives a rule for gradients; the condition says which point the curve passes through.

Answer. $dy/dx=2x+y$ , with $y(0)=3$ ; the initial gradient is $3$ .

Check. Substituting the starting point into the right-hand side gives the stated initial gradient without finding the whole curve.

Change only the sign of the $y$ -term in the slope law. The initial point stays fixed, but the initial direction changes.

Worked example 2: changing the slope law

Question. A curve passes through $(0,3)$ . Its gradient at $(x,y)$ is $2x-y$ . Write the equation and initial condition, then find the initial gradient.

Plan. Translate the new gradient expression and evaluate it at the same point.

\begin{aligned} \frac{dy}{dx}&=2x-y && \text{new rate law}\\ y(0)&=3 && \text{same initial point}\\ y'(0)&=2(0)-3=-3 && \text{initial rate} \end{aligned}

Why is the initial gradient negative? At the starting point, $2x=0$ and $-y=-3$ .

Answer. $dy/dx=2x-y$ , with $y(0)=3$ ; the initial gradient is $-3$ .

Check. Changing a sign in the law has not changed the condition $y(0)=3$ .

The same translation works in a physical setting. A tank of height $2.4\,\mathrm m$ , initially containing water to depth $0.3\,\mathrm m$ , has a depth rate proportional to the remaining height if

\frac{dh}{dt}=k(2.4-h),\quad k>0,\qquad h(0)=0.3

Here $t$ is in minutes, $h$ is in metres and $k$ has units $\mathrm{min^{-1}}$ . For $0\le h<2.4$ , the depth increases, more slowly as it approaches the top. The law is a modelling assumption, not a supplied formula.

Watch out

Wrong: “ $y'(0)=3$ ” as the translation of “the curve passes through $(0,3)$ ”. That describes a rate instead of a point. Right: $y(0)=3$ ; find any initial gradient from the differential equation.

Change only the coefficient of $y$ in example 2. Construct the equation; no solution for $y(x)$ is needed.

Population models and measured rates

A measured quantity and a measured rate can determine a proportionality constant before the differential equation is solved.

If population $N$ decreases at a rate proportional to $N$ , write $dN/dt=-kN$ , with $k>0$ . For $N=8000$ and a decrease of $120$ fish per year, $-120=-8000k$ , giving $k=0.015\,\mathrm{year^{-1}}$ . The calibrated law is $dN/dt=-0.015N$ , so at $N=4000$ it predicts a decrease of $60$ fish per year. A proportional decrease is not a fixed loss at every population.

Growth can depend on more than one factor. “Proportional both to population and to unused capacity” means multiply the factors, rather than add them.

A limited-growth rate law

For constant capacity $K$ and positive $k$ , the stated proportionality gives

\frac{dN}{dt}=kN(K-N)

For $0<N<K$ , growth is positive. At $N=0$ and $N=K$ , the rate is zero. If the model admits $N>K$ , its predicted rate is negative. Not in the formula booklet: learn it.

This is a continuous approximation to a count. Constant capacity and constant proportionality are assumptions; changing resources or migration can make the law unreliable.

Worked example 3: two proportional factors

Question. A lake has capacity $1000$ fish. Its growth rate is proportional to both $N$ and $1000-N$ . Initially $N=200$ , growing at $80$ fish per week. Construct and calibrate the equation, and predict the rate at $N=500$ .

Plan. Multiply the two factors, use the measured rate to find $k$ , then substitute the second population.

\begin{aligned} \frac{dN}{dt}&=kN(1000-N),\quad N(0)=200 && \text{law and condition}\\ 80&=k(200)(800) && \text{initial rate data}\\ k&=\frac1{2000} && \text{divide by }160000\\ \left.\frac{dN}{dt}\right|_{N=500}&=\frac{500(500)}{2000}=125 && \text{predict the rate} \end{aligned}

Why multiply $N$ and $1000-N$ ? The rate is proportional to both quantities together.

Answer. $dN/dt=N(1000-N)/2000$ , with $N(0)=200$ ; the predicted rate is $125$ fish per week. Here $k$ has units $\mathrm{fish^{-1}\,week^{-1}}$ .

Check. At $N=200$ , the calibrated law gives $200(800)/2000=80$ , as measured.

Change only the measured growth rate from $80$ to $160$ fish per week. This tests whether the fitted constant responds proportionally to the data.

Worked example 4: recalibrating the same model

Question. Use the same capacity and proportionality law, with $N(0)=200$ , but an initial growth rate of $160$ fish per week. Find the equation and the predicted rate at $N=500$ .

Plan. Recalculate $k$ from the new rate, leaving both population factors unchanged.

\begin{aligned} 160&=k(200)(800) && \text{new measured rate}\\ k&=\frac1{1000} && \text{calibrate}\\ \left.\frac{dN}{dt}\right|_{N=500}&=\frac{500(500)}{1000}=250 && \text{predict} \end{aligned}

Why has $k$ doubled? The population factors are unchanged, so doubling the measured rate doubles their multiplier.

Answer. $dN/dt=N(1000-N)/1000$ , with $N(0)=200$ ; the predicted rate is $250$ fish per week.

Check. The rate at $500$ has doubled from $125$ to $250$ , as has the calibration rate.

There is a useful surprise in the product law. Populations $200$ and $800$ have the same rate, since $200(800)=800(200)$ . More generally,

N(1000-N)=250000-(N-500)^2

so the largest rate occurs at half capacity, $N=500$ , without solving for $N(t)$ . This is a completed-square identity, not a supplied formula.

Watch out

Wrong: $dN/dt=k[N+(1000-N)]=1000k$ . Adding the proportional quantities turns the rate into a constant. Right: multiply them, $dN/dt=kN(1000-N)$ .

Change only the initial measured rate again. Check it back in your final equation.

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Pure Mathematics

Mechanics

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G6 - Constructing differential equations

Translating rate and proportionality language

Population models and measured rates

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