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Direct integration initial value problems

#035

Problem

y′=e−3x,y(0)=0y' = e^{-3x},\quad y(0)=0

Classification

  • exponential antiderivative
  • initial condition

Method

  • direct integration
  • initial value problem

Solution

y=−13e−3x+13y = -\frac{1}{3}e^{-3x} + \frac{1}{3}

Simulation & Code

Computational workSimulation
Graph visualization for entry #035
Download .py file
Python codeOpen
import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-2, 2, 800)

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

plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = -1/3e^(-3x) + 1/3", linewidth=2)
plt.plot(x, y_prime, label="y' = e^(-3x)", linewidth=2, color="red")
plt.title("Entry #035: 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 e^(-3x). After using integration, we can pinpoint the solution because the initial condition tells us y(0)=0.