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Posted on 19 Aug 2026Edited on 19 Aug 2026

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How to Use MATLAB for Mathematical Modelling

How to Use MATLAB for Mathematical Modelling

You could define those values like this:

javascript
Ta = 20;
T0 = 90;
k = 0.08;

tspan = [0 60];

The differential equation can then be written as an anonymous function:

javascript
coolingModel = @(t,T) -k*(T - Ta);

To calculate the temperature over time, use an ODE solver:

javascript
[t,T] = ode45(coolingModel,tspan,T0);

ode45 is one of MATLAB's general-purpose solvers for ordinary differential equations. The appropriate solver depends on the characteristics of the equation, so it should not automatically be treated as the answer for every ODE problem. MATLAB provides several other solvers for different numerical situations.

Now you have a set of calculated temperatures rather than just an equation on paper.

Plot the Results

The next step is to make those results easier to interpret.

javascript
plot(t,T,'LineWidth',2)
xlabel('Time (minutes)')
ylabel('Temperature (°C)')
title('Coffee Cooling Model')
grid on

A graph can tell you something that a column of numbers cannot.

You can immediately see the temperature falling quickly at first and then approaching the surrounding temperature more gradually.

This is one reason I find MATLAB useful for modelling. You are not restricted to calculating a final answer. You can experiment with the behaviour of the entire system.

If the graph looks completely unreasonable, that is also useful information. Perhaps the equation is wrong. Perhaps a parameter has been entered incorrectly. Perhaps the units do not match.

A strange graph is often the beginning of debugging rather than the end of the calculation.

Use Symbolic Mathematics When You Need More Than a Numerical Answer

Sometimes you do not just want a list of numbers. You want to work with the equation itself.

MATLAB's symbolic capabilities can help with tasks such as differentiation, integration, algebraic manipulation, and solving equations.

For example:

javascript
syms y(t)

ode = diff(y,t) == -0.08*(y - 20);

solution = dsolve(ode,y(0) == 90)

This gives you an analytical solution to the differential equation rather than only numerical values.

That can be extremely useful when checking a numerical model. If you can obtain an exact solution for a simple version of the problem, you can compare it with the result produced by your numerical solver.

I recommend doing this whenever practical. It gives you an independent check on your implementation.

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