Numerical solution of first-order differential equations

Numerical solution of first-order differential equations with Python and Tkinter today for you. I wanted to share that with you.

Differential equations code

import tkinter, math, typing


def ti(x, y):
    return 1 / (math.cos(x)) - y * math.tan(x)


# Euler's method
def fx(x_0, y_0, x_k, n_1) -> object:
    h = (x_k - x_0) / n_1
    for i in range(0, n_1):
        y1: typing.Union[float, typing.Any] = y_0 + h * ti(x_0, y_0)
        x1 = x_0 + h
        x_0 = x1
        y_0 = y1
    return y1


def rk(x_0, y_0, x_k, n_1):
    h = (x_k - x_0) / n_1
#Runge-Kutta 4th Order Method
    for i in range(0, n_1):
        k1 = h * ti(x_0, y_0)
        k2 = h * ti(x_0 + h / 2, y_0 + k1 / 2)
        k3 = h * ti(x_0 + h / 2, y_0 + k2 / 2)
        k4 = h * ti(x_0 + h, y_0 + k3)
        y1: typing.Union[float, typing.Any] = y_0 + (k1 + 2 * k2 + 2 * k3 + k4) / 6
        x_0 = x_0 + h
        y_0 = y1
    return y1


def calc_y1():
    x0 = float(x0_entry.get())
    y0 = float(y0_entry.get())
    xk = float(xk_entry.get())
    n = int(n_entry.get())
    try:
        y1: str = f'{fx(x0, y0, xk, n):.3f}'
        y2 = str = f'{rk(x0, y0, xk, n):.3f}'
    except:
        y1 = '?'
        y2 = '?'
    y1_label.configure(text=y1)
    y2_label.configure(text=y2)


root = tkinter.Tk()
frame = tkinter.Frame(root)
frame.pack()
t1_label = tkinter.Label(frame, bg='tan', text='Numerical solution of first-order differential equations',
                         font='arial 12')
t1_label.grid(row=0, column=0, columnspan=5, padx=25, pady=5)
x0_entry = tkinter.Entry(frame, width=10)
x0_entry.grid(row=1, column=2, pady=5)
x0_lebel = tkinter.Label(frame, text='Entry value of X: ')
x0_lebel.grid(row=1, column=1, pady=5)
y0_entry = tkinter.Entry(frame, width=10)
y0_entry.grid(row=2, column=2, pady=5)
y0_lebel = tkinter.Label(frame, text='Entry value of Y: ')
y0_lebel.grid(row=2, column=1, pady=5)
xk_entry = tkinter.Entry(frame, width=10)
xk_entry.grid(row=1, column=4, pady=5)
xk_lebel = tkinter.Label(frame, text='End value of Х: ')
xk_lebel.grid(row=1, column=3, pady=5)
n_entry = tkinter.Entry(frame, width=10)
n_entry.grid(row=2, column=4, pady=5)
n_lebel = tkinter.Label(frame, text='Number of partitions: ')
n_lebel.grid(row=2, column=3, pady=5)
y1_label = tkinter.Label(frame, text="?")
y1_label.grid(row=3, column=1, padx=5, pady=5)
y1_lebel = tkinter.Label(frame, text='Euler\'s method: ')
y1_lebel.grid(row=4, column=0, padx=5, pady=5)
y2_label = tkinter.Label(frame, text='?')
y2_label.grid(row=4, column=1, padx=5, pady=5)
y2_lebel = tkinter.Label(frame, text='Runge-Kutta 4th Order Method: ')
y2_lebel.grid(row=3, column=0, padx=5, pady=5)
eval_button = tkinter.Button(frame, bg='coral', text='Evaluate', width=10, command=calc_y1)
eval_button.grid(row=4, column=3, padx=5, pady=5)
exit_button = tkinter.Button(frame, bg='coral', text='Exit', width=10, command=root.destroy)
exit_button.grid(row=4, column=4, padx=5, pady=5)
root.mainloop()

There are two methods used: Euler’s and Runge-Kutta.
Enter data: X, Y, expected X and the number of partitions.
Click Evaluate to get the result.
After clicking Exit, the GUI will be closed down.