1. (5 points, List Answers) Use Euler's Method with step size \( \Delta t=0.1 \) to approximate \( y(0.1), y(0.2), y(0.3), \ldots, y(1) \) for the function \( y(t) \) satisfying the following initial value problem: \[ \frac{d y}{d t}=10(t+y), \quad y(0)=0.1 \] Please give exact answers for each of these ten values.

Let's apply Euler's method to approximate \( y(0.1), y(0.2), y(0.3), \ldots, y(1) \) using the given initial value problem. 1. **Step 0 (Initial Condition)**: \[ t_0 = 0, \quad y(0) = 0.1 \] 2. **Next steps (using Euler's method)**: We use the formula: \[ y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n) \] where \( f(t, y) = 10(t + y) \). - **Calculating each step**: For \( n = 0 \) (at \( t = 0 \)): \[ f(0, 0.1) = 10(0 + 0.1) = 1 \] \[ y(0.1) = 0.1 + 0.1 \cdot 1 = 0.1 + 0.1 = 0.2 \] For \( n = 1 \) (at \( t = 0.1 \)): \[ f(0.1, 0.2) = 10(0.1 + 0.2) = 3 \] \[ y(0.2) = 0.2 + 0.1 \cdot 3 = 0.2 + 0.3 = 0.5 \] For \( n = 2 \) (at \( t = 0.2 \)): \[ f(0.2, 0.5) = 10(0.2 + 0.5) = 7 \] \[ y(0.3) = 0.5 + 0.1 \cdot 7 = 0.5 + 0.7 = 1.2 \] For \( n = 3 \) (at \( t = 0.3 \)): \[ f(0.3, 1.2) = 10(0.3 + 1.2) = 15 \] \[ y(0.4) = 1.2 + 0.1 \cdot 15 = 1.2 + 1.5 = 2.7 \] For \( n = 4 \) (at \( t = 0.4 \)): \[ f(0.4, 2.7) = 10(0.4 + 2.7) = 31 \] \[ y(0.5) = 2.7 + 0.1 \cdot 31 = 2.7 + 3.1 = 5.8 \] For \( n = 5 \) (at \( t = 0.5 \)): \[ f(0.5, 5.8) = 10(0.5 + 5.8) = 66 \] \[ y(0.6) = 5.8 + 0.1 \cdot 66 = 5.8 + 6.6 = 12.4 \] For \( n = 6 \) (at \( t = 0.6 \)): \[ f(0.6, 12.4) = 10(0.6 + 12.4) = 126 \] \[ y(0.7) = 12.4 + 0.1 \cdot 126 = 12.4 + 12.6 = 25.0 \] For \( n = 7 \) (at \( t = 0.7 \)): \[ f(0.7, 25.0) = 10(0.7 + 25.0) = 257 \] \[ y(0.8) = 25.0 + 0.1 \cdot 257 = 25.0 + 25.7 = 50.7 \] For \( n = 8 \) (at \( t = 0.8 \)): \[ f(0.8, 50.7) = 10(0.8 + 50.7) = 517 \] \[