1. **State the problem:** We want to approximate the value of $y$ at $x=0.1$ for the differential equation $$\frac{dy}{dx} = x + y + xy$$ with initial condition $y(0) = 1$ using Euler's method with step size $h=0.025$. 2. **Recall Euler's method formula:** $$y_{n+1} = y_n + h f(x_n, y_n)$$ where $f(x,y) = x + y + xy$ is the derivative function. 3. **Calculate step by step:** - Initial values: $x_0=0$, $y_0=1$ - Step size: $h=0.025$ 4. **Step 1:** Calculate $f(x_0,y_0) = 0 + 1 + 0\times1 = 1$ $$y_1 = y_0 + h f(x_0,y_0) = 1 + 0.025 \times 1 = 1.025$$ $$x_1 = x_0 + h = 0 + 0.025 = 0.025$$ 5. **Step 2:** Calculate $f(x_1,y_1) = 0.025 + 1.025 + 0.025 \times 1.025 = 0.025 + 1.025 + 0.025625 = 1.075625$ $$y_2 = y_1 + h f(x_1,y_1) = 1.025 + 0.025 \times 1.075625 = 1.025 + 0.026890625 = 1.051890625$$ $$x_2 = 0.025 + 0.025 = 0.05$$ 6. **Step 3:** Calculate $f(x_2,y_2) = 0.05 + 1.051890625 + 0.05 \times 1.051890625 = 0.05 + 1.051890625 + 0.05259453125 = 1.15448515625$ $$y_3 = y_2 + h f(x_2,y_2) = 1.051890625 + 0.025 \times 1.15448515625 = 1.051890625 + 0.02886212890625 = 1.08075275390625$$ $$x_3 = 0.05 + 0.025 = 0.075$$ 7. **Step 4:** Calculate $f(x_3,y_3) = 0.075 + 1.08075275390625 + 0.075 \times 1.08075275390625 = 0.075 + 1.08075275390625 + 0.08105645654297 = 1.23680921044922$ $$y_4 = y_3 + h f(x_3,y_3) = 1.08075275390625 + 0.025 \times 1.23680921044922 = 1.08075275390625 + 0.03092023026123 = 1.11167298416748$$ $$x_4 = 0.075 + 0.025 = 0.1$$ 8. **Final answer:** The approximate value of $y$ at $x=0.1$ is $$\boxed{1.1117}$$ (rounded to 4 decimal places).