Use​ Euler's method to approximate the solution to the given initial value problem at the points xequals0.1​, 0.2​, 0.3​, 0.4​, and 0.5​, using steps of size 0.1 ​(hequals0.1​).
StartFraction dy Over dx EndFraction
equalsy left parenthesis 9 minus y right parenthesis​, ​y(0​)equals4

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.

Use Euler's method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1). dy/dx = y(9 - y), y(0) = 4.

Learn how to apply Euler's method to approximate the solution to the differential equation dy/dx = y(9 - y) with initial condition y(0) = 4. The solution covers the key steps at points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using step size h = 0.1. Suitable for college-level calculus students.

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