ODE solvers (ode45) in MATLAB - Step-by-Step Execution

Concept Flow - ODE solvers (ode45)

Define ODE function

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Set time span and initial value

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Call ode45 solver

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Solver runs steps internally

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Return time and solution arrays

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Plot or analyze results

The flow shows how you define the ODE, set initial conditions, call ode45, which solves step-by-step, then returns results.

Execution Sample

MATLAB

f = @(t,y) -2*y;
tspan = [0 5];
y0 = 1;
[t,y] = ode45(f,tspan,y0);
plot(t,y)

This code solves dy/dt = -2y from t=0 to 5 with y(0)=1 and plots the solution.

Execution Table

Step	t (time)	y (solution)	Action	Notes
1	0	1	Start at initial condition	Initial time and value
2	0.1	0.8187	Compute next y using ode45 step	Approximate y at t=0.1
3	0.2	0.6703	Compute next y	Approximate y at t=0.2
4	0.3	0.5488	Compute next y	Approximate y at t=0.3
5	0.4	0.4493	Compute next y	Approximate y at t=0.4
6	0.5	0.3679	Compute next y	Approximate y at t=0.5
...	...	...	Continue steps	Solver adapts step size internally
N	5	4.54e-05	Final step	Reached end of time span

💡 Reached t=5, end of time span for solution

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	...	Final
t	0	0.1	0.2	0.3	0.4	0.5	...	5
y	1	0.8187	0.6703	0.5488	0.4493	0.3679	...	4.54e-05

Key Moments - 3 Insights

Why does ode45 return arrays of t and y instead of just one value?

Because ode45 solves the ODE step-by-step over the time span, it returns arrays showing the solution at multiple time points (see execution_table rows 2 to N).

How does ode45 decide the size of each time step?

ode45 adapts the step size internally to keep the solution accurate, so time steps are not always equal (not shown explicitly but mentioned in execution_table notes).

What does the function handle @(t,y) -2*y represent?

It defines the ODE dy/dt = -2y, which ode45 uses to compute slopes at each step (see execution_sample code line 1).

Visual Quiz - 3 Questions

Look at the execution_table, what is the approximate value of y at step 3?

A 0.5488

B 0.8187

C 0.6703

D 0.4493

💡 Hint

Check the 'y (solution)' column at Step 3 in the execution_table.

At which step does the solver reach the end of the time span?

A Step N

B Step 5

C Step 6

D Step 1

💡 Hint

Look at the exit_note and the last row in execution_table.

If the initial value y0 was changed to 2, how would the variable_tracker change after Step 1?

A y would be about 1.0

B y would be about 1.6374 times 2

C y would be about 1.6374

D y would be about 1.6374 times 0.5

💡 Hint

Since y0 doubles, the values after Step 1 roughly double too (see variable_tracker y values).

Concept Snapshot

ode45 solves ODEs numerically in MATLAB.
Define the ODE as a function @(t,y).
Set time span and initial value.
Call [t,y] = ode45(odefun,tspan,y0).
 ode45 returns arrays of time and solution.
Use plot(t,y) to visualize results.

Full Transcript

This visual execution shows how MATLAB's ode45 solver works step-by-step. First, you define the ODE as a function of time and solution. Then you set the time range and initial value. Calling ode45 runs the solver, which calculates the solution at many time points by adapting step size for accuracy. The solver returns arrays of time and solution values. The example solves dy/dt = -2y from 0 to 5 starting at y=1. The execution table traces the solution values at each step, showing how y decreases exponentially. Key moments clarify why ode45 returns arrays, how step size adapts, and what the function handle means. The quiz tests understanding of solution values and solver behavior. The snapshot summarizes the main steps to use ode45 in MATLAB.