Solving ODEs (solve_ivp) in SciPy - Step-by-Step Execution

Concept Flow - Solving ODEs (solve_ivp)

Define ODE function f(t, y)

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Set initial conditions y0 and time span

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Call solve_ivp with f, time span, y0

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solve_ivp integrates step-by-step

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Collect solution times and values

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Use or plot solution results

We define the ODE, set initial values and time range, then call solve_ivp which calculates the solution step-by-step and returns the results.

Execution Sample

SciPy

from scipy.integrate import solve_ivp

def dydt(t, y):
    return -2 * y

sol = solve_ivp(dydt, [0, 5], [1])

This code solves the simple ODE dy/dt = -2y from time 0 to 5 with initial value y=1.

Execution Table

Step	t (time)	y (value)	Action	Notes
1	0.0	[1.0]	Start integration	Initial condition y(0)=1
2	0.1	[0.8187]	Compute next y	y(0.1) approx e^(-2*0.1) = 0.8187
3	0.2	[0.6703]	Compute next y	y(0.2) approx e^(-2*0.2) = 0.6703
4	0.3	[0.5488]	Compute next y	y(0.3) approx e^(-2*0.3) = 0.5488
5	0.4	[0.4493]	Compute next y	y(0.4) approx e^(-2*0.4) = 0.4493
6	0.5	[0.3679]	Compute next y	y(0.5) approx e^(-2*0.5) = 0.3679
7	5.0	[0.0000454]	Final step	y(5) approx e^(-2*5) = 4.54e-05

💡 Reached end of time span at t=5.0, integration complete.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
t	0.0	0.1	0.2	0.3	0.4	0.5	5.0
y	[1.0]	[0.8187]	[0.6703]	[0.5488]	[0.4493]	[0.3679]	[0.0000454]

Key Moments - 2 Insights

Why does the time t jump from 0.5 to 5.0 in the last step?

solve_ivp adapts step sizes for accuracy. It takes small steps at first, then jumps to the final time 5.0 to finish integration, as shown in the last row of the execution_table.

Why is y value decreasing at each step?

Because the ODE is dy/dt = -2y, which means y decreases exponentially over time. The execution_table rows show y values getting smaller as t increases.

Visual Quiz - 3 Questions

Look at the execution_table at Step 3. What is the approximate value of y?

A [0.8187]

B [0.5488]

C [0.6703]

D [0.3679]

💡 Hint

Check the 'y (value)' column at Step 3 in the execution_table.

At which step does the integration reach the final time t=5.0?

A Step 6

B Step 7

C Step 5

D Step 4

💡 Hint

Look at the 't (time)' column in the execution_table for the largest time value.

If the initial value y(0) was 2 instead of 1, how would y at Step 2 change?

A It would be about 1.6374

B It would stay 0.8187

C It would be about 0.3351

D It would be about 2.0

💡 Hint

Since y scales with initial value, double initial y doubles y at each step (see variable_tracker).

Concept Snapshot

solve_ivp(func, t_span, y0) solves ODE dy/dt=func(t,y)
- func(t,y) returns derivative
- t_span = [start, end] times
- y0 = initial value(s)
- Returns solution with times and y values
- Uses adaptive steps for accuracy

Full Transcript

We start by defining the ODE function that returns the derivative at any time and value. Then we set initial conditions and the time range to solve over. Calling solve_ivp runs the integration step-by-step, adapting step sizes for accuracy. The solution includes times and corresponding y values. For example, solving dy/dt = -2y from 0 to 5 with y(0)=1 shows y decreasing exponentially. The execution table traces each step's time and y value, showing how the solution progresses until the final time is reached.