Direct integration initial value problems

#039

Problem

y' = 2\csc x\cot x,\quad y(\pi/2)=1

Classification

trigonometric antiderivative
singularity
initial condition

Method

direct integration
initial value problem

Solution

y = -2\csc x + 3

Simulation & Code

Computational workSimulation

Download .py file

Python codeOpen

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-6, 6, 800)

with np.errstate(divide="ignore", invalid="ignore", over="ignore"):
    y = -2/np.sin(x) + 3
    y_prime = 2*np.cos(x)/(np.sin(x)**2)


def mask(values, limit=100):
    return np.where(np.isfinite(values) & (np.abs(values) <= limit), values, np.nan)

y = mask(y)
y_prime = mask(y_prime)
plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = -2csc(x) + 3", linewidth=2)
plt.plot(x, y_prime, label="y' = 2csc(x)cot(x)", linewidth=2, color="red")
plt.title("Entry #039: solution and derivative")
plt.xlabel("x")
plt.ylabel("value")
plt.grid(True, alpha=0.35)
plt.legend()
plt.tight_layout()
plt.show()

Handwritten derivation

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Takeaway & Interpretation

The differential equation tells us that the slope is 2csc(x)cot(x). After using integration, we can pinpoint the solution because the initial condition tells us y(pi/2)=1. This differential equation also has periodic singularities, like #038.