To solve the given problem using Euler's method, we'll calculate numerical estimates for the solution of the differential equation:

$y' = t + y, \quad y(0) = 1 \quad \text{on} \quad [0, 2], \quad \text{with step size} \, h = \frac{1}{2}.$

Step-by-step solution:

Euler's method is given by the iterative formula: $y_{n+1} = y_n + h \cdot f(t_n, y_n),$ where $f(t, y) = t + y$ in this case.

Initial Conditions:

• $t_0 = 0$, $y_0 = 1$
• Step size $h = 0.5$
• Compute $t_n = t_0 + nh$ for $n = 0, 1, 2, \dots$

Iteration 1:

At $t_0 = 0$, $y_0 = 1$: $f(t_0, y_0) = t_0 + y_0 = 0 + 1 = 1.$ $y_1 = y_0 + h \cdot f(t_0, y_0) = 1 + 0.5 \cdot 1 = 1.5.$ $t_1 = t_0 + h = 0.5.$

Iteration 2:

At $t_1 = 0.5$, $y_1 = 1.5$: $f(t_1, y_1) = t_1 + y_1 = 0.5 + 1.5 = 2.$ $y_2 = y_1 + h \cdot f(t_1, y_1) = 1.5 + 0.5 \cdot 2 = 2.5.$ $t_2 = t_1 + h = 1.0.$

Iteration 3:

At $t_2 = 1.0$, $y_2 = 2.5$: $f(t_2, y_2) = t_2 + y_2 = 1.0 + 2.5 = 3.5.$ $y_3 = y_2 + h \cdot f(t_2, y_2) = 2.5 + 0.5 \cdot 3.5 = 4.25.$ $t_3 = t_2 + h = 1.5.$

Iteration 4:

At $t_3 = 1.5$, $y_3 = 4.25$: $f(t_3, y_3) = t_3 + y_3 = 1.5 + 4.25 = 5.75.$ $y_4 = y_3 + h \cdot f(t_3, y_3) = 4.25 + 0.5 \cdot 5.75 = 7.125.$ $t_4 = t_3 + h = 2.0.$

Final Results:

• $t_0 = 0, \, y_0 = 1$
• $t_1 = 0.5, \, y_1 = 1.5$
• $t_2 = 1.0, \, y_2 = 2.5$
• $t_3 = 1.5, \, y_3 = 4.25$
• $t_4 = 2.0, \, y_4 = 7.125$

Approximation at $t = 2.0$:

$y(2.0) \approx 7.125$

Would you like further explanations or visualizations of this method? Here are some related questions:

1. How can we improve the accuracy of Euler's method?
2. Can we compare this result with the exact solution of the differential equation?
3. What are the limitations of Euler's method for stiff differential equations?
4. How does the step size affect the numerical approximation in Euler's method?
5. Can we apply higher-order methods (e.g., Runge-Kutta) to this problem?

Tip: Always verify numerical results by