Solve Differential Equations with ODEINT Function of SciPy module in Python

In science and engineering, many problems involve quantities that change over time like speed of a moving object or temperature of a cooling cup. These changes are often described using differential equations.

SciPy provides a function called odeint (from the scipy.integrate module) that helps solve these equations numerically. By giving it a function that describes how your system changes and some starting values, odeint calculates how the system behaves over time.

Syntax

scipy.integrate.odeint (func, y₀, t, args=())

Parameter:

• func: function that returns the derivative (dy/dt).
• y₀: initial conditions.
• t: time points to solve the ODE at.
• args (optional): extra values passed to func.

Solving Differential Equations

Let's solve an ordinary differential equation (ODE) using the odeint() function.

Example 1

\frac{dy}{dt} = -yt + 13

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

# Initial condition
y0 = 1

# Time points
t = np.linspace(0, 5)

# Define the differential equation using lambda
# dy/dt = -y * t + 13
dydt = lambda y, t: -y * t + 13

# Solve the ODE
y = odeint(dydt, y0, t)

plt.plot(t, y)
plt.xlabel("Time")
plt.ylabel("Y")
plt.show()

Output

Explanation:

• Defines and solves the ODE dy/dt = -y * t + 13 using odeint with initial value y = 1.
• The graph shows how y changes over time.
• y increases first then slows as -y * t term grows.

Example 2

\frac{dy}{dt} = 13e^t + y

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

# Initial condition
y0 = 1

# Time values
t = np.linspace(0, 5)

# Define ODE directly using lambda: dy/dt = 13 * e^t + y
dydt = lambda y, t: 13 * np.exp(t) + y

# Solve ODE
y = odeint(dydt, y0, t)

plt.plot(t, y)
plt.xlabel("Time")
plt.ylabel("Y")
plt.show()

Output

Explanation:

• Defines and solves ODE dy/dt = 13*eᵗ + y using odeint starting from y = 1.
• The graph shows y grows rapidly, driven by both exponential and linear terms.
• The exponential term accelerates the growth of y as time increases.

Example 3

\frac{dy}{dt} = \frac{1 - y}{1.95 - y} - \frac{y}{0.05 + y}

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

# Initial conditions
y0 = [0, 1, 2]

# Time values
t = np.linspace(1, 10)

# Define the ODE using lambda
dydt = lambda y, t: (1 - y) / (1.95 - y) - y / (0.05 + y)

# Solve the ODE
y = odeint(dydt, y0, t)

plt.plot(t, y)
plt.xlabel("Time")
plt.ylabel("Y")
plt.show()

Output

Explanation:

• Defines and solves ODE dy/dt = (1−y)/(1.95−y) − y/(0.05+y) using odeint with initial values y = 0, 1 and 2.
• The graph shows how y evolves over time for each initial value.
• The equation models a system with competing growth and decay rates, influenced by the values of y.

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