ODE solvers (ode45) in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of ode45 with a simple exponential decay

What is the output of the following MATLAB code snippet using ode45 to solve dy/dt = -2y with initial condition y(0) = 1 at t = 1?

MATLAB

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

A 0.1353

B 0.3679

C 0.5000

D 1.0000

💡 Hint

Recall the exact solution of dy/dt = -2y is y(t) = exp(-2t).

The exact solution at t=1 is exp(-2*1) = exp(-2) ≈ 0.1353. However, the code returns y(end) which corresponds to the last computed value by ode45 at t=1. The numerical solver approximates the solution and returns a value close to 0.1353. Option A matches this value. Option A (0.3679) corresponds to exp(-1), which is incorrect for this problem.

❓ Predict Output

intermediate

Number of output points from ode45 with specified time vector

Given the code below, how many points will the output y have?

MATLAB

odefun = @(t,y) y;
[t,y] = ode45(odefun, 0:0.5:2, 1);
length(y)

A 6

B 3

C 5

D 4

💡 Hint

When you provide a vector of time points to ode45, it returns solutions at those points.

The time vector 0:0.5:2 generates [0, 0.5, 1.0, 1.5, 2.0], which has 5 points. ode45 returns solution values at each of these points, so length(y) is 5.

❓ Predict Output

advanced

Output of ode45 solving a system of ODEs

What is the value of y(end,1) after running this code?

MATLAB

odefun = @(t,y) [y(2); -y(1)];
[t,y] = ode45(odefun, [0 pi/2], [0 1]);
y(end,1)

A 1.0000

B 0.0000

C -1.0000

D 0.7071

💡 Hint

This system models a simple harmonic oscillator with initial position 0 and velocity 1.

The system corresponds to the equations y1' = y2 and y2' = -y1, which describe sine and cosine functions. With initial conditions y1(0)=0 and y2(0)=1, the solution for y1(t) is sin(t). At t=pi/2, sin(pi/2) = 1. However, numerical approximation by ode45 may give a value close to 1 but slightly less. Option A (0.7071) corresponds to sin(pi/4), which is not the final time. Option A (1.0000) is the exact value. The numerical solver returns a value close to 1, so option A is correct.

❓ Predict Output

advanced

Error type when odefun returns wrong size output

What error does MATLAB produce when the function passed to ode45 returns a vector of incorrect size?

MATLAB

odefun = @(t,y) [y; y];
[t,y] = ode45(odefun, [0 1], 1);

A No error, code runs successfully.

B Index exceeds matrix dimensions.

C Undefined function or variable 'odefun'.

D Error using ode45: The function must return a column vector of the same length as y.

💡 Hint

ode45 expects the function to return a derivative vector matching the size of the initial condition.

ode45 requires the function to return a column vector with the same number of rows as the initial condition y. Returning a vector of different size causes an error message indicating the size mismatch.

🧠 Conceptual

expert

Behavior of ode45 with stiff equations

Which statement best describes the behavior of ode45 when solving stiff ordinary differential equations?

A ode45 is efficient and stable for stiff problems without any modifications.

B ode45 may require very small time steps and become inefficient for stiff problems.

C ode45 automatically switches to a stiff solver when stiffness is detected.

D ode45 cannot solve stiff problems and will always produce errors.

💡 Hint

Consider the nature of explicit Runge-Kutta methods like ode45 and their suitability for stiff problems.

ode45 is an explicit Runge-Kutta method designed for non-stiff problems. For stiff problems, it often requires very small time steps to maintain stability, making it inefficient. MATLAB provides other solvers like ode15s for stiff problems.

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