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Direct integration initial value problems

#034

Problem

y=2e2x,y(0)=3y' = 2e^{2x},\quad y(0)=3

Classification

  • exponential antiderivative
  • initial condition

Method

  • direct integration
  • initial value problem

Solution

y=e2x+2y = e^{2x} + 2

Simulation & Code

Computational workSimulation
Graph visualization for entry #034
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 = np.exp(2*x) + 2
    y_prime = 2*np.exp(2*x)

plt.figure(figsize=(8, 5))
plt.plot(x, y, label="y = e^(2x) + 2", linewidth=2)
plt.plot(x, y_prime, label="y' = 2e^(2x)", linewidth=2, color="red")
plt.title("Entry #034: 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 2e^(2x). After using integration, we can pinpoint the solution because the initial condition tells us y(0)=3.