Direct integration initial value problems

#042

Problem

y' = \frac{2x}{x^2 + 4},\quad y(0)=1

Classification

logarithmic antiderivative
initial condition

Method

direct integration
initial value problem
integration substitution

Solution

y = \ln|x^2 + 4| + 1 - \ln 4

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(x**2 + 4) + 1 - np.log(4)
    y_prime = 2*x/(x**2 + 4)

plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = ln(x^2 + 4) + 1 - ln(4)", linewidth=2)
plt.plot(x, y_prime, label="y' = 2x/(x^2 + 4)", linewidth=2, color="red")
plt.title("Entry #042: 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 2x/(x^2 + 4). After using integration, we can pinpoint the solution because the initial condition tells us y(0)=1.