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.