We are asked to use Euler's method to approximate the solution to the initial value problem:

$\frac{dy}{dx} = y(9 - y), \quad y(0) = 4$

Euler's Method Formula

Euler's method approximates the solution to a differential equation at discrete points using the formula:

$y_{n+1} = y_n + h \cdot f(x_n, y_n)$

Where:

• $y_n$ is the current approximation of $y$ at $x_n$,
• $f(x_n, y_n) = \frac{dy}{dx}$ is the function describing the derivative,
• $h$ is the step size.

We are given:

• Initial value: $y(0) = 4$,
• Step size $h = 0.1$,
• The differential equation $\frac{dy}{dx} = y(9 - y)$.

We'll compute $y$ for $x = 0.1, 0.2, 0.3, 0.4, 0.5$.

Step-by-Step Calculation

1. At $x = 0$, $y_0 = 4$: $f(0, 4) = 4(9 - 4) = 4 \times 5 = 20$ Euler's formula: $y_1 = 4 + 0.1 \times 20 = 4 + 2 = 6$ So, $y(0.1) \approx 6$.

2. At $x = 0.1$, $y_1 = 6$: $f(0.1, 6) = 6(9 - 6) = 6 \times 3 = 18$ Euler's formula: $y_2 = 6 + 0.1 \times 18 = 6 + 1.8 = 7.8$ So, $y(0.2) \approx 7.8$.

3. At $x = 0.2$, $y_2 = 7.8$: $f(0.2, 7.8) = 7.8(9 - 7.8) = 7.8 \times 1.2 = 9.36$ Euler's formula: $y_3 = 7.8 + 0.1 \times 9.36 = 7.8 + 0.936 = 8.736$ So, $y(0.3) \approx 8.736$.

4. At $x = 0.3$, $y_3 = 8.736$: $f(0.3, 8.736) = 8.736(9 - 8.736) = 8.736 \times 0.264 = 2.3059$ Euler's formula: $y_4 = 8.736 + 0.1 \times 2.3059 = 8.736 + 0.23059 = 8.96659$ So, $y(0.4) \approx 8.96659$.

5. At $x = 0.4$, $y_4 = 8.96659$: $f(0.4, 8.96659) = 8.96659(9 - 8.96659) = 8.96659 \times 0.03341 = 0.29965$ Euler's formula: $y_5 = 8.96659 + 0.1 \times 0.29965 = 8.96659 + 0.029965 = 8.996555$ So, $y(0.5) \approx 8.996555$.

Final Approximation

• $y(0.1) \approx 6$
• $y(0.2) \approx 7.8$
• $y(0.3) \approx 8.736$
• $y(0.4) \approx 8.96659$
• $y(0.5) \approx 8.996555$

Would you like more details on any step?

5 Relative Questions:

1. How does Euler's method differ from more accurate numerical methods like Runge-Kutta?
2. What is the effect of step size $h$ on the accuracy of Euler’s method?
3. Can Euler's method handle systems of differential equations, and if so, how?
4. How would the solution change if $h = 0.05$ were used instead of $h = 0.1$?
5. How is Euler’s method applied in real-world problems like population growth or physics?

Tip: Smaller step sizes in Euler's method generally yield more accurate results, but at the cost of more computation steps.