We will use Euler's method to approximate the solution to the differential equation:

$\frac{dy}{dx} = y(4 - y), \quad y(0) = 8$

with step size $h = 0.1$, at $x = 0.1, 0.2, 0.3, 0.4,$ and $0.5$.

Formula for Euler's Method:

$y_{n+1} = y_n + h \cdot f(x_n, y_n)$ where $f(x, y) = y(4 - y)$.

Step 1: Initialize

At $x_0 = 0$, $y_0 = 8$.

Step-by-Step Calculation

1. At $x_0 = 0, y_0 = 8$: $f(0, 8) = 8(4 - 8) = 8(-4) = -32$ $y_1 = y_0 + h \cdot f(0, 8) = 8 + 0.1(-32) = 8 - 3.2 = 4.8$

2. At $x_1 = 0.1, y_1 = 4.8$: $f(0.1, 4.8) = 4.8(4 - 4.8) = 4.8(-0.8) = -3.84$ $y_2 = y_1 + h \cdot f(0.1, 4.8) = 4.8 + 0.1(-3.84) = 4.8 - 0.384 = 4.416$

3. At $x_2 = 0.2, y_2 = 4.416$: $f(0.2, 4.416) = 4.416(4 - 4.416) = 4.416(-0.416) = -1.837056$ $y_3 = y_2 + h \cdot f(0.2, 4.416) = 4.416 + 0.1(-1.837056) = 4.416 - 0.1837056 = 4.2322944$

4. At $x_3 = 0.3, y_3 = 4.2322944$: $f(0.3, 4.2322944) = 4.2322944(4 - 4.2322944) = 4.2322944(-0.2322944) \approx -0.98297$ $y_4 = y_3 + h \cdot f(0.3, 4.2322944) = 4.2322944 + 0.1(-0.98297) = 4.2322944 - 0.098297 \approx 4.134$

5. At $x_4 = 0.4, y_4 = 4.134$: $f(0.4, 4.134) = 4.134(4 - 4.134) = 4.134(-0.134) \approx -0.554$ $y_5 = y_4 + h \cdot f(0.4, 4.134) = 4.134 + 0.1(-0.554) \approx 4.078$

Final Results

The approximate values of $y(x)$ are:

• $y(0.1) = 4.8$
• $y(0.2) \approx 4.416$
• $y(0.3) \approx 4.232$
• $y(0.4) \approx 4.134$
• $y(0.5) \approx 4.078$

Let me know if you need the calculations detailed further or for any clarifications!

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1. What is Euler's method and how does it approximate solutions?
2. Why is the step size important in Euler's method?
3. How does the accuracy of the method change with smaller step sizes?
4. What are the limitations of Euler's method?
5. Can Euler's method be applied to non-linear differential equations?

Tip: Always cross-check with smaller step sizes for higher accuracy when using Euler’s method.