Let's solve both parts using the **classical fourth-order Runge-Kutta (RK4) method** with a step size \( h = .5 \). --- ## **(a) Integrate \( f(x, y) = -2x + 12x^2 - 20x + 8.5 \) with \( y() = 1 \) and \( h = .5 \)** ### **Step 1: Compute RK4 increments (\( k_1, k_2, k_3, k_4 \)) for \( x_ = , y_ = 1 \)** RK4 formula: \[ \begin{align*} k_1 &= f(x_, y_) \\ k_2 &= f(x_ + \frac{h}{2}, y_ + \frac{h}{2} k_1) \\ k_3 &= f(x_ + \frac{h}{2}, y_ + \frac{h}{2} k_2) \\ k_4 &= f(x_ + h, y_ + h k_3) \\ y_1 &= y_ + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{align*} \] First, simplify \( f(x, y) \): \[ f(x, y) = -2x + 12x^2 - 20x + 8.5 = 12x^2 - 22x + 8.5 \] Plug in \( x_ = , y_ = 1 \): - \( k_1 = f(, 1) = 12()^2 - 22() + 8.5 = 8.5 \) - \( k_2 = f(.25, 1 + .25 \times 8.5) = f(.25, 3.125) = 12(.25)^2 - 22(.25) + 8.5 \) - \( 12 \times .0625 = .75 \) - \( -22 \times .25 = -5.5 \) - \( k_2 = .75 - 5.5 + 8.5 = 3.75 \) - \( k_3 = f(.25, 1 + .25 \times 3.75) = f(.25, 1.9375) = 3.75 \) (same as above) - \( k_4 = f(.5, 1 + .5 \times 3.75) = f(.5, 2.875) = 12(.5)^2 - 22(.5) + 8.5 \) - \( 12 \times .25 = 3 \) - \( -22 \times .5 = -11 \) - \( k_4 = 3 - 11 + 8.5 = .5 \) **Step 1 result:** \[ k_1 = 8.5,\quad k_2 = 3.75,\quad k_3 = 3.75,\quad k_4 = .5 \] --- ### **Step 2: Calculate \( y_1 \) at \( x_1 = .5 \)** \[ y_1 = y_ + \frac{.5}{6}(8.5 + 2 \times 3.75 + 2 \times 3.75 + .5) \] \[ = 1 + \frac{.5}{6}(8.5 + 7.5 + 7.5 + .5) = 1 + \frac{.5}{6}(24) = 1 + \frac{12}{6} = 1 + 2 = 3 \] **Final answer for (a):** \[ \boxed{y(.5) = 3} \] --- ## **(b) Integrate \( f(x, y) = 4e^{.8t} - .5y \) with \( y() = 2 \) and \( h = .5 \)** ### **Step 1: Compute RK4 increments (\( k_1, k_2, k_3, k_4 \)) for \( t_ = , y_ = 2 \)** - \( k_1 = f(, 2) = 4e^{} - .5 \times 2 = 4 - 1 = 3 \) - \( k_2 = f(.25, 2 + .25 \times 3) = f(.25, 2.75) = 4e^{.8 \times .25} - .5 \times 2.75 \) - \( e^{.2} \approx 1.2214 \) - \( 4 \times 1.2214 = 4.8856 \) - \( -.5 \times 2.75 = -1.375 \) - \( k_2 = 4.8856 - 1.375 = 3.5106 \) - \( k_3 = f(.25, 2 + .25 \times 3.5106) = f(.25, 2.8777) \) - \( -.5 \times 2.8777 = -1.43885 \) - \( k_3 = 4.8856 - 1.43885 = 3.44675 \) - \( k_4 = f(.5, 2 + .5 \times 3.44675) = f(.5, 3.723375) \) - \( e^{.4} \approx 1.4918 \) - \( 4 \times 1.4918 = 5.9672 \) - \( -.5 \times 3.723375 = -1.8616875 \) - \( k_4 = 5.9672 - 1.8616875 = 4.1055125 \) **Step 1 result:** \[ k_1 = 3,\quad k_2 \approx 3.5106,\quad k_3 \approx 3.44675,\quad k_4 \approx 4.10551 \] --- ### **Step 2: Calculate \( y_1 \) at \( t_1 = .5 \)** \[ y_1 = y_ + \frac{.5}{6}(3 + 2 \times 3.5106 + 2 \times 3.44675 + 4.10551) \] \[ = 2 + \frac{.5}{6}(3 + 7.0212 + 6.8935 + 4.10551) = 2 + \frac{.5}{6}(21.02021) = 2 + \frac{10.5101}{6} = 2 + 1.7517 \approx 3.75 \] **Final answer for (b):** \[ \boxed{y(.5) \approx 3.75} \] --- ## **Summary of Final Answers** - **(a) \( y(.5) = 3 \)** - **(b) \( y(.5) \approx 3.75 \)** Let me know if you want the calculations for further steps!