Direct integration initial value problems

#043

Problem

y' = \frac{1}{x + 2},\quad y(0)=0

Classification

logarithmic antiderivative
domain restriction
singularity
initial condition

Method

direct integration
initial value problem

Solution

y = \ln|x + 2| - \ln 2

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 = np.log(np.abs(x + 2)) - np.log(2)
    y_prime = 1/(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 = ln|x + 2| - ln(2)", linewidth=2)
plt.plot(x, y_prime, label="y' = 1/(x + 2)", linewidth=2, color="red")
plt.title("Entry #043: 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 1/(x + 2). After using integration, we can pinpoint the solution because the initial condition tells us y(0)=0. This differential equation has a shifted singularity at x=-2.