### Euler's Method Approximation
Euler's method is a numerical technique for solving differential equations. The formula for Euler's method is:
[tex]\[ y_{n+1} = y_n + \Delta x \cdot f(x_n, y_n) \][/tex]
Here, [tex]\( f(x, y) = x^3 \)[/tex] and we are given the initial condition [tex]\( y(0) = 5 \)[/tex], a step size [tex]\( \Delta x = 0.1 \)[/tex], and [tex]\( x \)[/tex] ranges up to 1.
Let's break this down step by step:
1. Initial Setup:
[tex]\[ x_0 = 0, \quad y_0 = 5 \][/tex]
2. Step 1:
- [tex]\( x_1 = x_0 + \Delta x = 0 + 0.1 = 0.1 \)[/tex]
- [tex]\( y_1 = y_0 + \Delta x \cdot (x_0^3) = 5 + 0.1 \cdot (0^3) = 5 \)[/tex]
3. Step 2:
- [tex]\( x_2 = x_1 + \Delta x = 0.1 + 0.1 = 0.2 \)[/tex]
- [tex]\( y_2 = y_1 + \Delta x \cdot (x_1^3) = 5 + 0.1 \cdot (0.1^3) = 5 + 0.1 \cdot 0.001 = 5.0001 \)[/tex]
4. Step 3:
- [tex]\( x_3 = x_2 + \Delta x = 0.2 + 0.1 = 0.3 \)[/tex]
- [tex]\( y_3 = y_2 + \Delta x \cdot (x_2^3) = 5.0001 + 0.1 \cdot (0.2^3) = 5.0001 + 0.1 \cdot 0.008 = 5.001 \)[/tex]
5. Continue in this fashion for each subsequent step.
After completing 10 steps (i.e., when [tex]\( x = 1 \)[/tex]), we get:
[tex]\[ y(1) \approx 5.2025 \][/tex]
Thus, the Euler approximation is:
[tex]\[ y(1) \approx 5.2025 \][/tex]
### Exact Solution
The exact solution of the differential equation can be found by solving the integral of the given differential equation.
Given:
[tex]\[ \frac{dy}{dx} = x^3 \][/tex]
Integrate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \int dy = \int x^3 \, dx \][/tex]
This gives us:
[tex]\[ y = \frac{x^4}{4} + C \][/tex]
Using the initial condition [tex]\( y(0) = 5 \)[/tex]:
[tex]\[ 5 = \frac{0^4}{4} + C \implies C = 5 \][/tex]
Therefore, the exact solution is:
[tex]\[ y = \frac{x^4}{4} + 5 \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ y(1) = \frac{1^4}{4} + 5 = \frac{1}{4} + 5 = 0.25 + 5 = 5.25 \][/tex]
So, the exact solution is:
[tex]\[ y(1) = 5.25 \][/tex]
### Conclusion
Summarizing the results:
- The Euler approximation at [tex]\( x = 1 \)[/tex] is [tex]\( y(1) \approx 5.2025 \)[/tex]
- The exact solution at [tex]\( x = 1 \)[/tex] is [tex]\( y(1) = 5.25 \)[/tex]
So, the solutions rounded to four decimal places are:
Euler approximation: [tex]\( y(1) \approx 5.2025 \)[/tex]
Exact solution: [tex]\( y(1) = 5.25 \)[/tex]
