P4.7 Differential equations
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Pure Mathematics 4 (A2), specification sections 5.2 and 6.4. Partial fractions are in P4.1, the integration methods in P4.5 and connected rates of change in P4.4; second-order equations and equations that do not separate are not part of this unit.
What this page covers
7 points
- What a differential equation is; general solutions (one arbitrary constant) and particular solutions
- Solving \(\frac{dy}{dx}=f(x)g(y)\) by separating the variables, including equations that give \(\ln|y|\)
- Separated integrals that need partial fractions, a standard result from the formulae booklet, or integration by parts
- Forming differential equations from statements about rates: proportional growth and decay
- Newton's law of cooling and draining tanks
- Limited (logistic) growth
- Finding constants from data, interpreting solutions, long-term behaviour and the limits of a model
In one minute
5 points
- A differential equation links a function and its derivative; the general solution of a first-order equation contains one arbitrary constant, and a given condition such as \(y=2\) when \(x=1\) fixes it to give a particular solution.
- Separate: \(\frac{dy}{dx}=f(x)g(y)\Rightarrow\int\frac1{g(y)}\,dy=\int f(x)\,dx\). Add one constant, on the \(x\) side, as soon as you integrate.
- Log constants: \(\ln|y|=f(x)+c\Rightarrow y=Ae^{f(x)}\), with \(A=\pm e^c\). Never \(y=e^{f(x)}+c\).
- Forming: "rate of change of \(N\)" is \(\frac{dN}{dt}\); "proportional to" brings in a constant \(k\); a decreasing quantity gets a minus sign with \(k>0\). Newton's law of cooling: \(\frac{d\theta}{dt}=-k(\theta-\theta_s)\).
- Interpret: use two pieces of data to find \(c\) and \(k\); for the long term let \(t\to\infty\) (\(e^{-kt}\to0\)); and say when the model stops being valid.
Contents
14 sections
- What is a differential equation, and what is a solution?
- How do you solve an equation by separating the variables?
- What happens to the constant when you get ln y?
- When do you need partial fractions?
- How do you form a differential equation from a description?
- How do you model cooling?
- How do you model a draining tank?
- How do you solve a limited-growth model?
- Exam tips
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What is a differential equation, and what is a solution?
Solving means integrating once, so the general solution contains one arbitrary constant. Each value of the constant gives a different curve: a whole family. A boundary or initial condition ("\(y=2\) when \(x=1\)", "initially the temperature is 85 °C") picks out one member, the particular solution.
The equation itself tells you the gradient at every point. Drawing a short segment with that gradient at many points gives a slope field, and every solution curve follows the segments. To verify a solution, differentiate it (implicitly if it is not in the form \(y=\dots\)) and show that the result matches the equation.
(a) Show that \(y^3=3x^3+C\), where \(C\) is a constant, satisfies the differential equation \(\frac{dy}{dx}=\frac{3x^2}{y^2}\). (b) Find the particular solution for which \(y=2\) when \(x=1\). (c) Use your answer to (b) to find \(y\) when \(x=2\), giving your answer to 3 significant figures.
(a) Differentiate implicitly: \(3y^2\frac{dy}{dx}=9x^2\), so \(\frac{dy}{dx}=\frac{9x^2}{3y^2}=\frac{3x^2}{y^2}\) ✓.
(b) Substitute \(x=1\), \(y=2\): \(8=3+C\), so \(C=5\) and \(y^3=3x^3+5\).
(c) \(y^3=24+5=29\), so \(y=\sqrt[3]{29}=3.0723\ldots=\) 3.07.
How do you solve an equation by separating the variables?
An equation is separable if the right-hand side factorises into a function of \(x\) times a function of \(y\). Then each variable can be moved to its own side and integrated on its own.
Factorise first: \(\frac{dy}{dx}=xy+x=x(y+1)\) is separable. \(\frac{dy}{dx}=x+y\) is not, and cannot be solved this way.
Given that \(\frac{dy}{dx}=(2x+1)e^{-y}\) and that \(y=0\) when \(x=0\), (a) find \(y\) in terms of \(x\); (b) find the exact value of \(y\) when \(x=2\), and give it to 3 significant figures.
(a) \(e^{-y}=1\div e^y\), so multiply both sides by \(e^y\): \(\int e^y\,dy=\int(2x+1)\,dx\), giving \(e^y=x^2+x+c\). Use \(x=0\), \(y=0\): \(e^0=1=c\). So \(e^y=x^2+x+1\) and \(y=\ln(x^2+x+1)\).
(b) \(y=\ln(4+2+1)=\) \(\ln7=1.9459\ldots=1.95\).
\(x^2+x+1=\left(x+\frac12\right)^2+\frac34>0\), so the logarithm is defined for every \(x\).
Taking ln of each term separately: from \(e^y=x^2+x+1\), writing \(y=\ln x^2+\ln x+\ln1\). The logarithm applies to the whole right-hand side: \(y=\ln(x^2+x+1)\).
What happens to the constant when you get ln y?
Whenever \(g(y)=y\) (or \(y+b\)), the \(y\) side integrates to \(\ln|y|\). Exponentiate the whole right-hand side: \(\ln|y|=f(x)+c\Rightarrow|y|=e^{f(x)+c}=e^c\,e^{f(x)}\). Replace \(\pm e^c\) by a single constant \(A\): \(y=Ae^{f(x)}\). Equivalently, write the constant as \(\ln A\) from the start: \(\ln y=f(x)+\ln A\). In context questions the quantity is usually positive (a mass, a population), so the modulus can be dropped; say so.
The variables \(x\) and \(y\) satisfy \(\frac{dy}{dx}=y\sin x\), with \(y>0\), and \(y=3\) when \(x=0\). (a) Show that \(y=3e^{1-\cos x}\). (b) Find the greatest value of \(y\), giving the exact value and the value to 3 significant figures.
(a) Separate: \(\int\frac1y\,dy=\int\sin x\,dx\), so \(\ln y=-\cos x+c\) (\(y>0\), so no modulus). Exponentiate the whole right-hand side: \(y=e^ce^{-\cos x}=Ae^{-\cos x}\). Use \(x=0\), \(y=3\) (\(\cos0=1\)): \(3=Ae^{-1}\), so \(A=3e\) and \(y=3e\times e^{-\cos x}=3e^{1-\cos x}\) ✓.
(b) The power \(1-\cos x\) is greatest when \(\cos x=-1\): \(y_{\max}=3e^2\approx22.2\) (at \(x=\pi,3\pi,\dots\)).
Writing \(\ln y=-\cos x+c\) as \(y=e^{-\cos x}+c\). The constant is part of the power, so it becomes a multiplier: \(y=Ae^{-\cos x}\). Putting \(x=0\) into the wrong form gives a different, incorrect curve.
When do you need partial fractions?
After separating, either side may be a rational function such as \(\frac2{x(x+2)}\) or \(\frac1{P(6-P)}\). Split it into partial fractions (P4.1) so each piece integrates to a logarithm. Watch the sign: \(\int\frac1{6-P}\,dP=-\ln|6-P|\), because the derivative of \(6-P\) is \(-1\). Collect the logarithms into one with the laws of logs, then exponentiate: \(\ln y=\ln\frac{x}{x+2}+c\Rightarrow y=\frac{Ax}{x+2}\). Other separated integrals may need a standard result from the formulae booklet (\(\int\tan x\,dx=\ln|\sec x|\)) or integration by parts (\(\int xe^x\,dx\)): see P4.5.
(a) Express \(\frac2{x(x+2)}\) in partial fractions. (b) Given that \(\frac{dy}{dx}=\frac{2y}{x(x+2)}\) for \(x>0\), \(y>0\), and that \(y=6\) when \(x=1\), find \(y\) in terms of \(x\). (c) Describe what happens to \(y\) as \(x\) becomes large.
(a) \(2\equiv P(x+2)+Qx\); \(x=0\) gives \(P=1\), \(x=-2\) gives \(Q=-1\): \(\frac1x-\frac1{x+2}\).
(b) Separate and use (a): \(\int\frac1y\,dy=\int\left(\frac1x-\frac1{x+2}\right)dx\), so \(\ln y=\ln x-\ln(x+2)+\ln A\). Combine: \(\ln y=\ln\frac{Ax}{x+2}\), so \(y=\frac{Ax}{x+2}\). With \(x=1\), \(y=6\): \(6=\frac A3\), \(A=18\), so \(y=\dfrac{18x}{x+2}\).
(c) Divide top and bottom by \(x\): \(y=\frac{18}{1+2/x}\to18\) as \(x\to\infty\). \(y\) increases towards 18 but never reaches it.
Writing the constant as \(\ln A\) when every other term is a logarithm lets you combine everything into one log in a single line. Either form of the constant is correct; choose the one that makes the rearranging shortest.
How do you form a differential equation from a description?
Identify the quantity and the time variable: "the rate of change of the mass \(m\)" is \(\frac{dm}{dt}\). "Proportional to" gives a constant: proportional to \(m\) is \(km\); inversely proportional to \(m\) is \(\frac km\); proportional to the square root of \(h\) is \(k\sqrt h\). If the quantity decreases, write a minus sign and state \(k>0\), so that \(\frac{dm}{dt}=-km\) is negative whenever \(m>0\). "Proportional to the difference between \(\theta\) and 20" gives \(k(\theta-20)\); "to the product of \(P\) and \(6-P\)" gives \(kP(6-P)\). When the rate is given for a different quantity (volume, when the question wants depth), connect them with the chain rule first (P4.4).
| Description | Differential equation | Shape of the solution |
|---|---|---|
| \(N\) increases at a rate proportional to \(N\) | \(\frac{dN}{dt}=kN\), \(k>0\) | \(N=N_0e^{kt}\): exponential growth |
| \(m\) decreases at a rate proportional to \(m\) | \(\frac{dm}{dt}=-km\), \(k>0\) | \(m=m_0e^{-kt}\): decay towards 0 |
| temperature \(\theta\) falls at a rate proportional to its excess over the surroundings at \(\theta_s\) | \(\frac{d\theta}{dt}=-k(\theta-\theta_s)\) | \(\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}\): levels off at \(\theta_s\) |
| depth \(h\) falls at a rate proportional to \(\sqrt h\) | \(\frac{dh}{dt}=-k\sqrt h\) | \(\sqrt h\) falls linearly: the tank empties in a finite time |
| \(P\) grows at a rate proportional to \(P(L-P)\) | \(\frac{dP}{dt}=kP(L-P)\) | S-shaped curve levelling off at \(L\) |
A drug is injected into a patient's bloodstream. The mass of the drug, \(m\) mg, decreases at a rate proportional to the mass present. Initially there is 40 mg; 3 hours later there is 25 mg. (a) Write down a differential equation for \(m\) in terms of the time \(t\) hours. (b) Show that \(m=40e^{-kt}\), where \(k\) is a positive constant. (c) Show that \(k=\frac13\ln1.6\). (d) Find the time at which the mass has fallen to 5 mg. (e) Find the time taken for the mass to halve.
(a) Decreasing, so a minus sign with \(k>0\): \(\dfrac{dm}{dt}=-km\).
(b) Separate (\(m>0\)): \(\int\frac1m\,dm=\int-k\,dt\), so \(\ln m=-kt+c\) and \(m=Ae^{-kt}\). At \(t=0\), \(m=40\), so \(A=40\): \(m=40e^{-kt}\) ✓.
(c) Use \(t=3\), \(m=25\): \(e^{-3k}=\frac{25}{40}\), so \(e^{3k}=1.6\) and \(k=\frac13\ln1.6\) ✓ (\(=0.1567\ldots\)).
(d) \(40e^{-kt}=5\Rightarrow e^{kt}=8\Rightarrow t=\frac{\ln8}k=\frac{3\ln8}{\ln1.6}=13.273\ldots=\) 13.3 hours.
(e) \(e^{kt}=2\Rightarrow t=\frac{3\ln2}{\ln1.6}=\) 4.42 hours, the same from any starting mass.
How do you model cooling?
A hot body (\(\theta>\theta_s\)) cools; a cold body (\(\theta<\theta_s\)) warms up; the same equation covers both. Separate with \(\theta-\theta_s\) as one block: \(\int\frac1{\theta-\theta_s}\,d\theta=\ln|\theta-\theta_s|\). The temperature approaches \(\theta_s\) but never reaches it in the model, so a "long-term" answer is \(\theta_s\).
A bowl of soup at 85 °C is placed in a room at a constant 20 °C. The temperature of the soup, \(\theta\) °C, \(t\) minutes later satisfies \(\frac{d\theta}{dt}=-k(\theta-20)\), where \(k\) is a positive constant. After 10 minutes the temperature is 60 °C. (a) Show that \(\theta=20+65e^{-kt}\). (b) Find the exact value of \(k\). (c) Find the time at which the soup reaches 35 °C. (d) State what happens to \(\theta\) in the long term.
(a) Separate (\(\theta>20\) throughout): \(\int\frac1{\theta-20}\,d\theta=\int-k\,dt\), so \(\ln(\theta-20)=-kt+c\) and \(\theta-20=Ae^{-kt}\). At \(t=0\), \(A=85-20=65\): \(\theta=20+65e^{-kt}\) ✓.
(b) \(t=10\), \(\theta=60\): \(65e^{-10k}=40\Rightarrow e^{10k}=\frac{13}8\Rightarrow\) \(k=\frac1{10}\ln\frac{13}8\) (\(=0.04855\ldots\)).
(c) \(65e^{-kt}=15\Rightarrow e^{kt}=\frac{13}3\Rightarrow t=\frac{10\ln(13/3)}{\ln(13/8)}=30.20\ldots=\) 30.2 minutes.
(d) \(e^{-kt}\to0\) as \(t\to\infty\), so \(\theta\to20\): the soup cools towards room temperature but stays above it.
Using \(\theta=85e^{-kt}\). That model cools towards 0 °C, not to room temperature. The exponential describes the excess temperature \(\theta-20\), which starts at 65, not 85.
How do you model a draining tank?
Water flowing out through a hole near the base leaves faster when the water is deeper. A common model: the depth falls at a rate proportional to \(\sqrt h\), so \(\frac{dh}{dt}=-k\sqrt h\). Separate with a power: \(\int h^{-1/2}\,dh=2\sqrt h\). The solution has \(\sqrt h\) falling linearly, so, unlike exponential decay, the tank empties in a finite time. The formula for \(h\) is a squared bracket, which would start rising again after the tank is empty, so state the interval of validity.
Water leaks from a tank through a small hole in its base. The depth of water is \(h\) metres at time \(t\) minutes, and the depth decreases at a rate proportional to the square root of the depth. Initially the depth is 1.44 m; after 10 minutes it is 1.00 m. (a) Write down a differential equation for \(h\). (b) Find the general solution. (c) Find the values of the constants, and hence express \(h\) in terms of \(t\). (d) Find the time taken for the tank to empty. (e) State, with a reason, the interval of \(t\) for which your formula is valid.
(a) \(\dfrac{dh}{dt}=-k\sqrt h\), \(k>0\).
(b) Separate: \(\int h^{-1/2}\,dh=\int-k\,dt\), so \(2\sqrt h=-kt+c\).
(c) \(t=0\), \(h=1.44\): \(c=2\times1.2=2.4\). \(t=10\), \(h=1\): \(2=-10k+2.4\), so \(k=0.04\). Then \(\sqrt h=1.2-0.02t\) and \(h=(1.2-0.02t)^2\).
(d) \(h=0\): \(1.2-0.02t=0\Rightarrow\) \(t=60\) minutes.
(e) \(0\le t\le60\): after 60 minutes the tank is empty, but \((1.2-0.02t)^2\) would increase again.
With two unknown constants (\(c\) and \(k\)), use the initial condition first: at \(t=0\) the \(kt\) term vanishes and \(c\) comes out alone. Then the second data point gives \(k\).
How do you solve a limited-growth model?
Unlimited growth \(\frac{dP}{dt}=kP\) is unrealistic for a population with limited food or space. Multiplying by \((L-P)\) makes the growth rate fall to zero as \(P\) approaches the limit \(L\). Separating gives \(\int\frac1{P(L-P)}\,dP\): split into partial fractions, integrate to logs (watch the minus sign on \(\ln|L-P|\)), combine and exponentiate. Make \(P\) the subject by collecting the \(P\) terms on one side and factorising. The answer usually comes out in the form \(\frac{L}{1+Be^{-ct}}\), which shows \(P\to L\) at once.
The population of fish in a lake, \(P\) hundred, \(t\) years after the lake is stocked, is modelled by \(\frac{dP}{dt}=\frac{P(6-P)}{15}\), \(0<P<6\). Initially there are 100 fish. (a) Express \(\frac6{P(6-P)}\) in partial fractions. (b) Show that \(\ln\left(\frac{5P}{6-P}\right)=0.4t\). (c) Hence show that \(P=\frac6{1+5e^{-0.4t}}\). (d) Find the time taken for the population to reach 400 fish. (e) State the long-term size of the population according to the model.
(a) \(6\equiv a(6-P)+bP\); \(P=0\) gives \(a=1\), \(P=6\) gives \(b=1\): \(\frac1P+\frac1{6-P}\).
(b) Separate: \(\frac{15}{P(6-P)}=\frac52\times\frac6{P(6-P)}\), so \(\int\frac52\left(\frac1P+\frac1{6-P}\right)dP=\int1\,dt\). The second term gives \(-\ln(6-P)\): \(\frac52\big[\ln P-\ln(6-P)\big]=t+c\), so \(\ln\frac P{6-P}=0.4t+c'\). At \(t=0\), \(P=1\) (100 fish): \(c'=\ln\frac15\). So \(\ln\frac P{6-P}-\ln\frac15=0.4t\), i.e. \(\ln\left(\frac{5P}{6-P}\right)=0.4t\) ✓.
(c) Remove the log: \(5P=(6-P)e^{0.4t}\), so \(P(5+e^{0.4t})=6e^{0.4t}\) and \(P=\frac{6e^{0.4t}}{5+e^{0.4t}}=\frac6{1+5e^{-0.4t}}\) ✓ (divide top and bottom by \(e^{0.4t}\)).
(d) 400 fish means \(P=4\): \(1+5e^{-0.4t}=1.5\Rightarrow e^{-0.4t}=0.1\Rightarrow t=2.5\ln10=\) 5.76 years.
(e) \(e^{-0.4t}\to0\), so \(P\to6\): 600 fish.
Integrating \(\frac1{6-P}\) as \(+\ln(6-P)\). The derivative of \(6-P\) is \(-1\), so the integral is \(-\ln(6-P)\). With the wrong sign the solution has no limiting value and part (e) cannot come out as 6.
Exam tips for P4.7
Write a line of the form \(\int(\text{only }y)\,dy=\int(\text{only }x)\,dx\) with both integral signs showing. Factorise the right-hand side first if needed: \(xy-2x=x(y-2)\).
Add \(+c\) on the line where the integral signs disappear. Adding it after rearranging (for example after exponentiating) puts it in the wrong place, and the particular solution that follows is wrong.
Leave \(k\) as \(\frac13\ln1.6\) or \(\frac1{10}\ln\frac{13}8\) and use the exact value (or the full calculator value) in later parts. Rounding \(k\) to 2 significant figures early can move a final answer by a whole unit in the third figure.
For a decreasing quantity the equation must make the derivative negative. After solving, check: does the formula give the initial value at \(t=0\)? Does it move in the right direction? State answers with units and in the words of the question ("600 fish", "30.2 minutes").
For long-term behaviour let \(t\to\infty\) and say which term tends to zero. For a comment on the model, link it to the context: a population cannot be a fraction of a fish, a tank cannot have negative depth, room temperature might not stay constant.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\ln y=2x+c\) gives \(y=e^{2x}+c\). | \(e^{2x+c}=e^ce^{2x}\): the constant multiplies. | \(y=Ae^{2x}\). |
| Each side needs its own constant. | Two constants combine into one, \(c_2-c_1\). | One constant, on one side, is enough for a first-order equation. |
| \(\int\frac1{6-P}\,dP=\ln(6-P)\). | The inner derivative of \(6-P\) is \(-1\). | \(-\ln|6-P|+c\). |
| "Decreases at a rate proportional to \(m\)" is \(\frac{dm}{dt}=km\). | With \(k>0\) this makes \(m\) increase. | \(\frac{dm}{dt}=-km\), \(k>0\). |
| Newton's law of cooling gives \(\theta=\theta_0e^{-kt}\). | That cools towards 0 °C. | \(\theta-\theta_s\) decays: \(\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}\). |
| A solution formula holds for all \(t\ge0\). | The context can stop the model, e.g. an empty tank. | State the interval of validity (\(0\le t\le60\) in worked example 7). |
How this topic connects
P4.1, P4.4 and P4.5
The separated integrals use partial fractions (P4.1) and substitution or integration by parts (P4.5); connected rates of change (P4.4) turn a rate for volume into a rate for depth before you separate.
P3.5 Exponential and logarithmic functions
\(e^x\), \(\ln x\), \(\int\frac1x\,dx=\ln|x|\) and the exponential growth and decay models of P3 are exactly the solutions of \(\frac{dN}{dt}=\pm kN\).
Mechanics and IAL Physics
Variable acceleration links displacement, velocity and acceleration by differentiation and integration with respect to time, the same "rate of change" language. Radioactive decay (\(\frac{dN}{dt}=-\lambda N\)), capacitor discharge and half-life are proportional-decay differential equations with exponential solutions.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{dy}{dx}=f(x)g(y)\Rightarrow\int\frac1{g(y)}\,dy=\int f(x)\,dx\) | separating the variables | No — learn it |
| \(\int\frac1x\,dx=\ln|x|+c\); \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c\) | the \(y\) side of most model equations | No — learn it |
| \(\ln|y|=f(x)+c\Rightarrow y=Ae^{f(x)}\) | removing the logarithm | No — derive it |
| \(\frac{dN}{dt}=kN\Rightarrow N=N_0e^{kt}\) | proportional growth (\(k>0\)) or decay (\(k<0\)) | No — derive it |
| \(\frac{d\theta}{dt}=-k(\theta-\theta_s)\Rightarrow\theta=\theta_s+(\theta_0-\theta_s)e^{-kt}\) | Newton's law of cooling | No — derive it |
| \(\frac{dh}{dt}=-k\sqrt h\Rightarrow2\sqrt h=-kt+c\) | draining tank | No — derive it |
| \(\int\tan x\,dx=\ln|\sec x|+c\); \(\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx\) | harder separated integrals | Yes |
Check your understanding
1. How many arbitrary constants does the general solution of a first-order differential equation contain, and how is it removed?
One. A given pair of values (an initial or boundary condition) fixes it and gives the particular solution.
2. Solve \(\frac{dy}{dx}=2y\), given that \(y=5\) when \(x=0\).
\(\int\frac1y\,dy=\int2\,dx\), so \(\ln|y|=2x+c\) and \(y=Ae^{2x}\). With \(y=5\) at \(x=0\), \(A=5\): \(y=5e^{2x}\).
3. The volume \(V\) of a melting block of ice decreases at a rate proportional to \(V^{2/3}\). Write down a differential equation.
\(\frac{dV}{dt}=-kV^{2/3}\), where \(k\) is a positive constant (minus because \(V\) decreases).
4. A model gives \(\theta=15+50e^{-0.1t}\). What are the initial and long-term temperatures?
At \(t=0\), \(\theta=15+50=65\); as \(t\to\infty\), \(e^{-0.1t}\to0\), so \(\theta\to15\), the temperature of the surroundings.
5. Is \(\frac{dy}{dx}=xy-3x\) separable?
Yes: \(xy-3x=x(y-3)\), so \(\int\frac1{y-3}\,dy=\int x\,dx\), giving \(y=3+Ae^{x^2/2}\).
Specification checklist
- (P4 6.4) I can recognise a separable first-order differential equation, factorising the right-hand side if necessary, and separate the variables.
- (P4 6.4) I can integrate both sides to find a general solution with one arbitrary constant, including writing \(\ln|y|=f(x)+c\) as \(y=Ae^{f(x)}\).
- (P4 6.4) I can use a given condition to find a particular solution and, when asked, write \(y\) explicitly in terms of \(x\).
- (P4 6.4) I can use partial fractions, standard integrals and integration by parts on either side of a separated equation.
- (P4 5.2) I can form a first-order differential equation from a description of a rate of change, using a constant of proportionality with the correct sign.
- (P4 5.2) I can find the constants in a model from data, interpret the solution in context, find long-term behaviour and comment on when the model is valid.