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.

Sure, let's apply **Euler's Method** with a step size \( \Delta t = 0.1 \) to approximate the values of \( y(t) \) for \( t = 0.1, 0.2, \ldots, 1.0 \) given the initial value problem: \[ \frac{dy}{dt} = 10(t + y), \quad y(0) = 0.1 \] **Euler's Method Formula:** \[ y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n) \] where \( f(t, y) = 10(t + y) \). Let's compute each step iteratively: 1. **Initial Condition:** \[ t_0 = 0.0, \quad y_0 = 0.1 \] 2. **Step-by-Step Calculations:** \[ \begin{array}{|c|c|c|c|} \hline n & t_n & y_n & \text{Computation} \\ \hline 0 & 0.0 & 0.1 & \text{Given} \\ \hline 1 & 0.1 & y_1 = y_0 + 0.1 \cdot 10(0.0 + 0.1) = 0.1 + 0.1 = 0.2 \\ \hline 2 & 0.2 & y_2 = y_1 + 0.1 \cdot 10(0.1 + 0.2) = 0.2 + 0.3 = 0.5 \\ \hline 3 & 0.3 & y_3 = y_2 + 0.1 \cdot 10(0.2 + 0.5) = 0.5 + 0.7 = 1.2 \\ \hline 4 & 0.4 & y_4 = y_3 + 0.1 \cdot 10(0.3 + 1.2) = 1.2 + 1.5 = 2.7 \\ \hline 5 & 0.5 & y_5 = y_4 + 0.1 \cdot 10(0.4 + 2.7) = 2.7 + 3.1 = 5.8 \\ \hline 6 & 0.6 & y_6 = y_5 + 0.1 \cdot 10(0.5 + 5.8) = 5.8 + 6.3 = 12.1 \\ \hline 7 & 0.7 & y_7 = y_6 + 0.1 \cdot 10(0.6 + 12.1) = 12.1 + 12.7 = 24.8 \\ \hline 8 & 0.8 & y_8 = y_7 + 0.1 \cdot 10(0.7 + 24.8) = 24.8 + 25.5 = 50.3 \\ \hline 9 & 0.9 & y_9 = y_8 + 0.1 \cdot 10(0.8 + 50.3) = 50.3 + 51.1 = 101.4 \\ \hline 10 & 1.0 & y_{10} = y_9 + 0.1 \cdot 10(0.9 + 101.4) = 101.4 + 102.3 = 203.7 \\ \hline \end{array} \] **Final Approximations:** \[ \begin{aligned} y(0.1) &= 0.2 \\ y(0.2) &= 0.5 \\ y(0.3) &= 1.2 \\ y(0.4) &= 2.7 \\ y(0.5) &= 5.8 \\ y(0.6) &= 12.1 \\ y(0.7) &= 24.8 \\ y(0.8) &= 50.3 \\ y(0.9) &= 101.4 \\ y(1.0) &= 203.7 \\ \end{aligned} \] These values are the Euler's Method approximations of \( y(t) \) at each specified time step from \( t = 0.1 \) to \( t = 1.0 \).