To solve this problem using Euler's method, we proceed as follows:

Given Data

1. Differential equation: $\frac{dy}{dx} = 8x^3$
2. Initial condition: $y(0) = 3$
3. Step size $h = 0.1$
4. We want to approximate $y(0.5)$.

Euler's Method Formula

Euler's method for approximating the solution to the differential equation $\frac{dy}{dx} = f(x, y)$ with initial condition $y(x_0) = y_0$ is: $y_{n+1} = y_n + h \cdot f(x_n, y_n)$ where $x_{n+1} = x_n + h$.

Step-by-Step Approximation

Using the formula above and $f(x, y) = 8x^3$:

1. Starting Point: $x_0 = 0$, $y_0 = 3$

2. Step 1 ($x = 0.1$):

   • $f(x_0, y_0) = 8 \cdot (0)^3 = 0$
   • $y_1 = y_0 + h \cdot f(x_0, y_0) = 3 + 0.1 \cdot 0 = 3$

3. Step 2 ($x = 0.2$):

   • $f(x_1, y_1) = 8 \cdot (0.1)^3 = 0.008$
   • $y_2 = y_1 + h \cdot f(x_1, y_1) = 3 + 0.1 \cdot 0.008 = 3.0008$

4. Step 3 ($x = 0.3$):

   • $f(x_2, y_2) = 8 \cdot (0.2)^3 = 0.064$
   • $y_3 = y_2 + h \cdot f(x_2, y_2) = 3.0008 + 0.1 \cdot 0.064 = 3.0072$

5. Step 4 ($x = 0.4$):

   • $f(x_3, y_3) = 8 \cdot (0.3)^3 = 0.216$
   • $y_4 = y_3 + h \cdot f(x_3, y_3) = 3.0072 + 0.1 \cdot 0.216 = 3.0288$

6. Step 5 ($x = 0.5$):

   • $f(x_4, y_4) = 8 \cdot (0.4)^3 = 0.512$
   • $y_5 = y_4 + h \cdot f(x_4, y_4) = 3.0288 + 0.1 \cdot 0.512 = 3.08$

Answer

The approximate value of $y(0.5)$ using Euler's method with $h = 0.1$ is 3.080 (rounded to three decimal places).

Would you like further details on this solution or the exact approach?

Here are some related questions for further practice:

1. How would the answer change if the step size $h$ were smaller?
2. Can you find the exact solution to this differential equation?
3. How does Euler’s method approximate solutions to differential equations?
4. What is the impact of step size $h$ on the accuracy of Euler's method?
5. Could we use other numerical methods to approximate $y(0.5)$, and how would they compare?

Tip: Euler’s method is simple but can be inaccurate with larger step sizes; smaller $h$ values improve precision.