ODE solvers (ode45) in MATLAB

Introduction

ODE solvers like ode45 help you find solutions to equations that describe how things change over time. They do the hard math for you.

When you want to model how a population grows or shrinks over time.

When you need to simulate the motion of a falling object with air resistance.

When you want to understand how a chemical reaction changes concentration over time.

When you want to predict the temperature change in a cooling cup of coffee.

When you want to solve physics problems involving changing speeds or positions.

Syntax

MATLAB

tspan = [t0 tf];
[t, y] = ode45(@odefun, tspan, y0);

function dydt = odefun(t, y)
    dydt = ...; % define the derivative here
end

tspan is the time interval you want to solve over, from start t0 to end tf.

odefun is a function that returns the rate of change (derivative) at time t and state y.

Examples

This solves the simple decay equation dy/dt = -2y from time 0 to 5 starting at y=1.

MATLAB

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

This solves a pendulum motion where y(1) is angle and y(2) is angular velocity.

MATLAB

function dydt = pendulum(t, y)
    g = 9.81; L = 1;
    dydt = [y(2); -(g/L)*sin(y(1))];
end

[t, y] = ode45(@pendulum, [0 10], [pi/4; 0]);

Sample Program

This program solves the equation dy/dt = -2y and shows the values of y at different times.

MATLAB

function simple_decay_example
    tspan = [0 5];
    y0 = 1;
    [t, y] = ode45(@(t,y) -2*y, tspan, y0);
    disp(table(t, y))
end

Important Notes

ode45 is good for many problems but not all. It works best for smooth problems without sudden jumps.

You can plot the results using plot(t, y) to see how the solution changes over time.

Summary

ode45 solves equations that describe change over time.

You give it a function for the rate of change, a time range, and a starting value.

It returns time points and the solution values at those points.

Practice

1. What does the MATLAB function ode45 primarily do?

easy

A. It solves ordinary differential equations numerically.

B. It plots graphs of functions automatically.

C. It calculates the derivative of a function symbolically.

D. It integrates functions analytically.

Solution

1. Step 1: Understand the purpose of ode45

   The ode45 function in MATLAB is designed to solve ordinary differential equations (ODEs) using a numerical method called Runge-Kutta.

2. Step 2: Compare options with ode45's function

   Options A, C, and D describe plotting, symbolic differentiation, and analytical integration, which are not what ode45 does.

3. Final Answer:

   It solves ordinary differential equations numerically. -> Option A

4. Quick Check:

   ode45 solves ODEs numerically = B [OK]

Hint: Remember ode45 solves ODEs numerically, not symbolically or graphing [OK]

Common Mistakes:

• Confusing ode45 with plotting functions
• Thinking ode45 does symbolic math
• Assuming ode45 integrates analytically

2. Which of the following is the correct syntax to call ode45 for solving dy/dt = -2y with initial value y(0)=1 over time interval [0 5]?

easy

A. [t,y] = ode45(@(y,t) -2*y, [0 5], 1);

B. [t,y] = ode45(@(t,y) -2*y, [0 5], 1);

C. [t,y] = ode45(@(t,y) 2*y, 0:5, 1);

D. [t,y] = ode45(@(t,y) -2*y, 0, 5, 1);

Solution

1. Step 1: Check function handle and argument order

   The function handle for ode45 must be @(t,y) for time t and variable y. [t,y] = ode45(@(t,y) -2*y, [0 5], 1); uses @(t,y) correctly with the expression -2*y.

2. Step 2: Verify time span and initial value format

   Time span must be a vector [0 5], and initial value is a scalar 1. [t,y] = ode45(@(t,y) -2*y, [0 5], 1); matches this format exactly.

3. Final Answer:

   [t,y] = ode45(@(t,y) -2*y, [0 5], 1); -> Option B

4. Quick Check:

   Correct syntax: @(t,y), time vector, initial value = A [OK]

Hint: Use @(t,y) for function and time vector in brackets [OK]

Common Mistakes:

• Swapping t and y in function handle
• Using colon operator instead of vector for time span
• Passing time points as separate arguments

3. Consider the code:

f = @(t,y) t - y;
[t,y] = ode45(f, [0 2], 1);
disp(y(end));

What is the output approximately?

medium

A. 1.3

B. 1.0

C. 0.5

D. 2.0

Solution

1. Step 1: Understand the ODE and initial condition

   The ODE is dy/dt = t - y with y(0) = 1, solved over [0 2]. This is a linear ODE where y tends to follow t but is pulled down by -y.

2. Step 2: Estimate solution behavior at t=2

   Numerical solution with ode45 typically gives y around 1.3 at t=2 for this ODE and initial value.

3. Final Answer:

   1.3 -> Option A

4. Quick Check:

   ode45 output y(end) ≈ 1.3 for dy/dt = t - y [OK]

Hint: Recall ode45 returns solution array; last value is y at final time [OK]

Common Mistakes:

• Assuming output equals initial value
• Confusing function input order
• Ignoring numerical approximation

4. Identify the error in this code snippet:

f = @(y,t) y^2 - t;
[t,y] = ode45(f, [0 1], 0);
plot(t,y);

medium

A. Time span must be a column vector.

B. Initial value cannot be zero.

C. Function handle arguments are in wrong order.

D. Plot command is missing labels.

Solution

1. Step 1: Check function handle argument order

   ODE functions for ode45 must have arguments in order (t,y), but here it is @(y,t), which is incorrect.

2. Step 2: Verify other parts of the code

   Initial value zero is allowed, time span as [0 1] is valid, and plot labels are optional, so these are not errors.

3. Final Answer:

   Function handle arguments are in wrong order. -> Option C

4. Quick Check:

   Function handle must be @(t,y), not @(y,t) [OK]

Hint: Always use @(t,y) for ode45 function handles [OK]

Common Mistakes:

• Swapping t and y in function handle
• Thinking initial value can't be zero
• Believing time span must be column vector

5. You want to solve the system:

dx/dt = y
dy/dt = -x

with initial conditions x(0)=0, y(0)=1 over [0 10]. Which MATLAB code correctly uses ode45 to solve this system?

hard

A. f = @(t,u) [u(1); u(2)];
[t,u] = ode45(f, [0 10], [0 1]);

B. f = @(t,x,y) [y; -x];
[t,u] = ode45(f, [0 10], [0 1]);

C. f = @(u,t) [u(2); -u(1)];
[t,u] = ode45(f, [0 10], [0 1]);

D. f = @(t,u) [u(2); -u(1)];
[t,u] = ode45(f, [0 10], [0 1]);

Solution

1. Step 1: Define function handle for system of ODEs

   The system has two variables x and y, combined as vector u where u(1)=x and u(2)=y. The function must be @(t,u) returning [y; -x], which is [u(2); -u(1)].

2. Step 2: Check initial conditions and call to ode45

   Initial conditions are [0 1] for x=0 and y=1. Time span is [0 10]. f = @(t,u) [u(2); -u(1)]; [t,u] = ode45(f, [0 10], [0 1]); matches this exactly.

3. Final Answer:

   f = @(t,u) [u(2); -u(1)]; [t,u] = ode45(f, [0 10], [0 1]); -> Option D

4. Quick Check:

   System ODEs use @(t,u) with vector u = D [OK]

Hint: Use vector input @(t,u) and return derivatives as vector [OK]

Common Mistakes:

• Using separate arguments for variables in function handle
• Swapping order of arguments in function handle
• Returning wrong derivative vector