Solving ODEs (solve_ivp) in SciPy - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the function solve_ivp in SciPy do?

solve_ivp solves ordinary differential equations (ODEs) numerically. It finds the values of variables over time given their rates of change.

beginner

What are the main inputs required by solve_ivp?

The main inputs are:

• A function defining the ODE system (derivatives)
• The time interval to solve over
• Initial values of the variables

beginner

How do you define the ODE system function for solve_ivp?

The function takes time t and current values y, and returns the derivatives as an array or list.

Example: def f(t, y): return [y[1], -y[0]]

beginner

What does the output of solve_ivp contain?

The output is an object with:

• t: time points where solution is computed
• y: solution values at those times
• Other info like success status

beginner

Why is solve_ivp useful in real life?

It helps model things that change over time, like population growth, chemical reactions, or physics problems, when exact formulas are hard to find.

What does the function passed to solve_ivp represent?

A The derivatives of variables with respect to time

B The initial values of variables

C The final solution values

D The time points to solve at

The function defines how variables change over time, i.e., their derivatives.

Which of these is NOT an input to solve_ivp?

A Initial values

B Time interval

C Plot style

D Function defining derivatives

Plot style is not an input to solve_ivp. It only needs the ODE function, time span, and initial values.

What does solve_ivp return?

A The derivative function

B Only the final value of the solution

C A plot of the solution

D An object with time points and solution values

The output contains arrays of time points and corresponding solution values.

If you want to solve an ODE from time 0 to 10 with initial value 1, which is correct?

A <code>solve_ivp(f, [0, 10], [1])</code>

B <code>solve_ivp(f, 10, 1)</code>

C <code>solve_ivp([0, 10], f, [1])</code>

D <code>solve_ivp(f, [1], [0, 10])</code>

The correct syntax is solve_ivp(function, time_span, initial_values).

Why might you use solve_ivp instead of solving ODEs by hand?

A Because it always gives exact answers

B Because many ODEs have no simple exact solution

C Because it is faster to write code than math

D Because it plots the solution automatically

Many ODEs are too complex for exact solutions, so numerical methods like solve_ivp help find approximate answers.

Explain how to use solve_ivp to solve a simple ODE step-by-step.

Think about what you need to tell the solver and what you get back.

Describe a real-world example where solving an ODE with solve_ivp would be helpful.

Consider things like population, physics, or chemistry.

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