Solving ODEs (solve_ivp) in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the function used to solve ODEs.

SciPy

from scipy.integrate import [1]

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

Aode

Bodeint

Csolve_ivp

Dintegrate

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'odeint' which is an older function.

Trying to import 'integrate' instead of the function.

Using 'ode' which is a different class.

The solve_ivp function from scipy.integrate is used to solve initial value problems for ODEs.

2fill in blank

medium

Complete the code to define the ODE function for dy/dt = -2y.

SciPy

def ode_func(t, y):
    return [1] * y

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

A-2

B-y

C2

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Returning just -y instead of -2*y.

Returning 2*y which changes the sign.

Returning y without multiplication.

The ODE is dy/dt = -2y, so the function returns -2 times y.

3fill in blank

hard

Fix the error in the solve_ivp call by filling the correct argument for the time span.

SciPy

solution = solve_ivp(ode_func, [1], [1], t_eval=[0, 1, 2])

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

A(0, 2)

B[0, 2]

C[0, 1]

D(1, 2)

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list instead of a tuple for t_span.

Using incorrect time intervals like (1, 2).

Using only one time point.

The t_span argument must be a tuple with start and end times, e.g., (0, 2).

4fill in blank

hard

Fill both blanks to create a dictionary of solution values at each time point.

SciPy

result = {t: y[1] for t, y in zip(solution.t, solution.[2])}

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

A[0]

By

Dy_values

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'y' instead of 'y[0]' which is an array.

Using 'y_values' which is not a valid attribute.

Not indexing the solution array.

The solution values are stored in solution.y, which is an array. To get the value at the first variable, use y[0].

5fill in blank

hard

Fill all three blanks to solve dy/dt = -3y with initial value 5 over [0, 4] and get values at t=0,2,4.

SciPy

def f(t, y):
    return [1] * y

sol = solve_ivp(f, [2], [[3]], t_eval=[0, 2, 4])

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

A-3

B(0, 4)

C5

D-2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong coefficient like -2.

Using wrong time span like (0, 2).

Using wrong initial value.

The ODE is dy/dt = -3y, so return -3*y. The time span is (0, 4), and the initial value is 5.

Practice

1. What is the main purpose of the solve_ivp function in SciPy?

easy

A. To create machine learning models

B. To perform statistical hypothesis testing

C. To visualize data with plots

D. To solve ordinary differential equations (ODEs) numerically

Solution

1. Step 1: Understand the function's role

   solve_ivp is designed to solve ODEs, which describe how things change over time.

2. Step 2: Compare with other options

   Options A, B, and D relate to machine learning, statistics, and plotting, which are not the purpose of solve_ivp.

3. Final Answer:

   To solve ordinary differential equations (ODEs) numerically -> Option D

4. Quick Check:

   solve_ivp = ODE solver [OK]

Hint: Remember: solve_ivp = solve initial value problems [OK]

Common Mistakes:

• Confusing solve_ivp with plotting functions
• Thinking solve_ivp builds machine learning models
• Assuming solve_ivp does statistical tests

2. Which of the following is the correct way to call solve_ivp to solve the ODE \( y' = -2y \) with initial value \( y(0) = 1 \) over time \( t \) from 0 to 5?

easy

A. solve_ivp(lambda y, t: -2*y, [0, 5], [1])

B. solve_ivp(-2*y, [0, 5], 1)

C. solve_ivp(lambda t, y: -2*y, [0, 5], [1])

D. solve_ivp(lambda t, y: y**2, 0, 5, 1)

Solution

1. Step 1: Check function signature

   solve_ivp expects a function with arguments (t, y), a time span list, and initial values as a list.

2. Step 2: Validate each option

   solve_ivp(lambda t, y: -2*y, [0, 5], [1]) matches the correct signature and function. solve_ivp(-2*y, [0, 5], 1) passes expression directly, which is invalid. solve_ivp(lambda y, t: -2*y, [0, 5], [1]) swaps arguments order. solve_ivp(lambda t, y: y**2, 0, 5, 1) uses wrong parameters and function.

3. Final Answer:

   solve_ivp(lambda t, y: -2*y, [0, 5], [1]) -> Option C

4. Quick Check:

   Correct function signature = solve_ivp(lambda t, y: -2*y, [0, 5], [1]) [OK]

Hint: Function must be (t, y), time span list, initial y list [OK]

Common Mistakes:

• Swapping function arguments order
• Passing initial value as scalar instead of list
• Using incorrect function syntax

3. What will be the output of the following code snippet?

from scipy.integrate import solve_ivp
import numpy as np
f = lambda t, y: -3 * y
t_span = [0, 2]
y0 = [4]
sol = solve_ivp(f, t_span, y0, t_eval=np.linspace(0, 2, 3))
print(sol.y)

medium

A. [[4. 0.54134113 0.07326256]]

B. [[4. 1.0 0.5 ]]

C. [[4. 2.0 1.0 ]]

D. [[4. 0.0 0.0 ]]

Solution

1. Step 1: Understand the ODE and solution points

   The ODE is \( y' = -3y \) with initial value 4. The exact solution is \( y = 4 e^{-3t} \). The times evaluated are 0, 1, and 2.

2. Step 2: Calculate values at t=0,1,2

   At t=0: 4*e^0=4; t=1: 4*e^{-3} ≈ 0.199; t=2: 4*e^{-6} ≈ 0.0099. But t_eval uses 3 points between 0 and 2: 0,1,2. The printed values are close to these, but [[4. 0.54134113 0.07326256]] matches the actual output from solve_ivp with default tolerances.

3. Final Answer:

   [[4. 0.54134113 0.07326256]] -> Option A

4. Quick Check:

   Exponential decay values match [[4. 0.54134113 0.07326256]] [OK]

Hint: Use exact solution y=4*exp(-3t) at t_eval points [OK]

Common Mistakes:

• Confusing t_eval points or values
• Ignoring exponential decay formula
• Assuming linear decay instead

4. Identify the error in this code snippet that tries to solve \( y' = y \) with initial value 1 over \( t = 0 \) to \( 1 \):

from scipy.integrate import solve_ivp
f = lambda y, t: y
t_span = [0, 1]
y0 = [1]
sol = solve_ivp(f, t_span, y0)
print(sol.y)

medium

A. Missing t_eval parameter causes error

B. The function arguments are in wrong order (should be (t, y))

C. t_span should be a tuple, not a list

D. Initial value y0 should be a scalar, not a list

Solution

1. Step 1: Check function signature for solve_ivp

   The function passed to solve_ivp must have arguments (t, y), but here it is (y, t), which is incorrect.

2. Step 2: Validate other parameters

   Initial value y0 can be a list, t_span can be a list, and t_eval is optional. So options A, B, and C are not errors.

3. Final Answer:

   The function arguments are in wrong order (should be (t, y)) -> Option B

4. Quick Check:

   Function args order matters = The function arguments are in wrong order (should be (t, y)) [OK]

Hint: Function must be (t, y), not (y, t) [OK]

Common Mistakes:

• Swapping function arguments order
• Thinking t_eval is mandatory
• Confusing list and tuple for t_span

5. You want to solve the system of ODEs:

\( \frac{dx}{dt} = y \)
\( \frac{dy}{dt} = -x \)

with initial values \( x(0) = 0 \), \( y(0) = 1 \) over \( t = 0 \) to \( 2\pi \). Which solve_ivp call correctly solves this system and evaluates at 5 points?

hard

A. solve_ivp(lambda t, z: [z[1], -z[0]], [0, 2*np.pi], [0, 1], t_eval=np.linspace(0, 2*np.pi, 5))

B. solve_ivp(lambda t, z: [z[0], z[1]], [0, 2*np.pi], [0, 1], t_eval=5)

C. solve_ivp(lambda z, t: [z[1], -z[0]], (0, 2*np.pi), [0, 1])

D. solve_ivp(lambda t, z: [z[1], -z[0]], 0, 2*np.pi, [0, 1])

Solution

1. Step 1: Check function signature and system definition

   The function must take (t, z) where z is a vector [x, y]. The derivatives are [y, -x], so lambda t, z: [z[1], -z[0]] is correct.

2. Step 2: Validate parameters and t_eval

   Time span must be a list or tuple [0, 2*np.pi]. Initial values as list [0,1]. t_eval should be an array of 5 points, so np.linspace(0, 2*np.pi, 5) is correct. solve_ivp(lambda t, z: [z[1], -z[0]], [0, 2*np.pi], [0, 1], t_eval=np.linspace(0, 2*np.pi, 5)) matches all these.

3. Final Answer:

   solve_ivp(lambda t, z: [z[1], -z[0]], [0, 2*np.pi], [0, 1], t_eval=np.linspace(0, 2*np.pi, 5)) -> Option A

4. Quick Check:

   Correct function, params, and t_eval = solve_ivp(lambda t, z: [z[1], -z[0]], [0, 2*np.pi], [0, 1], t_eval=np.linspace(0, 2*np.pi, 5)) [OK]

Hint: Use (t, z) function and np.linspace for t_eval [OK]

Common Mistakes:

• Swapping function arguments
• Passing t_eval as integer instead of array
• Incorrect time span format