ODE solvers (ode45) in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to solve the ODE dy/dt = -2y using ode45.

MATLAB

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

Drag options to blanks, or click blank then click option'

A-2*y

B2*y

Cy-2

Dt*y

Attempts:

3 left

💡 Hint

Common Mistakes

Using '2*y' instead of '-2*y' changes the sign and the solution behavior.

Writing 'y-2' or 't*y' does not represent the given ODE.

The function inside ode45 defines the derivative dy/dt = -2*y.

2fill in blank

medium

Complete the code to plot the solution y versus time t.

MATLAB

plot(t, [1]);
xlabel('Time'); ylabel('Solution y');

Drag options to blanks, or click blank then click option'

Ay0

Bt

Ctspan

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Plotting t against t or tspan does not show the solution.

Using y0 plots only the initial value, not the solution.

We plot the solution values y against time t.

3fill in blank

hard

Fix the error in the function definition for ode45 to solve dy/dt = y^2 - t.

MATLAB

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

Drag options to blanks, or click blank then click option'

Ay^2 - y

By**2 - t

Cy^2 - t

Dt - y^2

Attempts:

3 left

💡 Hint

Common Mistakes

Using '**' causes syntax errors in MATLAB.

Changing the equation changes the problem.

In MATLAB, '^' is used for power, so 'y^2 - t' is correct.

4fill in blank

hard

Fill both blanks to create a function handle for dy/dt = sin(t) - y.

MATLAB

odefun = @(t, y) [1] - [2];
[t, y] = ode45(odefun, [0 10], 1);

Drag options to blanks, or click blank then click option'

Asin(t)

Bcos(t)

Cy

Dt

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'cos(t)' instead of 'sin(t)' changes the equation.

Using 't' instead of 'y' changes the dependent variable.

The derivative is dy/dt = sin(t) - y, so the function is 'sin(t) - y'.

5fill in blank

hard

Fill all three blanks to solve dy/dt = -3y + 2t and plot the result.

MATLAB

odefun = @(t, y) [1] * [2] + 2 * [3];
[t, y] = ode45(odefun, [0 4], 0);
plot(t, y);

Drag options to blanks, or click blank then click option'

A-3

By

Ct

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up variables y and t.

Using wrong coefficients changes the equation.

The function is dy/dt = -3*y + 2*t, so blanks are '-3', 'y', and 't'.

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