Direct integration initial value problems

#046

Problem

y' = x^2e^{x^3},\quad y(0)=0

Classification

exponential antiderivative
growth
initial condition

Method

direct integration
initial value problem
integration substitution

Solution

y = \frac{1}{3}e^{x^3} - \frac{1}{3}

Simulation & Code

Computational workSimulation

Download .py file

Python codeOpen

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-1.5, 1.5, 800)

with np.errstate(divide="ignore", invalid="ignore", over="ignore"):
    y = (1/3)*np.exp(x**3) - 1/3
    y_prime = x**2*np.exp(x**3)

plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = (1/3)e^(x^3) - 1/3", linewidth=2)
plt.plot(x, y_prime, label="y' = x^2e^(x^3)", linewidth=2, color="red")
plt.title("Entry #046: 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 x^2e^(x^3). After using integration, we can pinpoint the solution because the initial condition tells us y(0)=0.