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Direct integration initial value problems

#050

Problem

y′=xx2+1,y(0)=2y' = \frac{x}{\sqrt{x^2 + 1}},\quad y(0)=2

Classification

  • power rule
  • initial condition

Method

  • direct integration
  • initial value problem
  • integration substitution

Solution

y=x2+1+1y = \sqrt{x^2 + 1} + 1

Simulation & Code

Computational workSimulation
Graph visualization for entry #050
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.sqrt(x**2 + 1) + 1
    y_prime = x/np.sqrt(x**2 + 1)

plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = sqrt(x^2 + 1) + 1", linewidth=2)
plt.plot(x, y_prime, label="y' = x/sqrt(x^2 + 1)", linewidth=2, color="red")
plt.title("Entry #050: 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 slope is controlled by x/sqrt(x^2+1). It is negative for x<0, zero at x=0, and positive for x>0, so the solution decreases into x=0 and increases after it. The IVP selects one specific curve with a smooth minimum at x=0.