1. **Problem Statement:** Solve the initial value problem (IVP) given by the differential equation $$xy' = e^{4x} \tan(y) + 9y$$ with initial condition $y(0.6) = 1$ using the Runge-Kutta 4th order (RK4) method to approximate $y(1)$ with step size $h=0.2$. 2. **Rewrite the equation:** We want to express $y'$ explicitly: $$y' = \frac{e^{4x} \tan(y) + 9y}{x}$$ Note: Since $x=0.6$ initially, division by $x$ is valid. 3. **RK4 method formula:** Given $y_{n}$ at $x_n$, the next value $y_{n+1}$ at $x_{n+1} = x_n + h$ is $$ \begin{aligned} k_1 &= f(x_n, y_n) \\ k_2 &= f\left(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_1\right) \\ k_3 &= f\left(x_n + \frac{h}{2}, y_n + \frac{h}{2}k_2\right) \\ k_4 &= f(x_n + h, y_n + hk_3) \\ y_{n+1} &= y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{aligned} $$ where $f(x,y) = \frac{e^{4x} \tan(y) + 9y}{x}$. 4. **Step 1: From $x_0=0.6$, $y_0=1$ to $x_1=0.8$** - Calculate $k_1$: $$k_1 = f(0.6,1) = \frac{e^{4\times0.6} \tan(1) + 9\times1}{0.6}$$ Calculate values: $e^{2.4} \approx 11.023$, $\tan(1) \approx 1.5574$ $$k_1 = \frac{11.023 \times 1.5574 + 9}{0.6} = \frac{17.17 + 9}{0.6} = \frac{26.17}{0.6} = 43.62$$ - Calculate $k_2$: $$k_2 = f\left(0.7, 1 + 0.1 \times 43.62\right) = f(0.7, 5.362)$$ Calculate $f(0.7,5.362)$: $e^{2.8} \approx 16.444$, $\tan(5.362) \approx \tan(5.362 - 2\pi) = \tan(-0.921) \approx -1.312$ $$k_2 = \frac{16.444 \times (-1.312) + 9 \times 5.362}{0.7} = \frac{-21.58 + 48.26}{0.7} = \frac{26.68}{0.7} = 38.11$$ - Calculate $k_3$: $$k_3 = f\left(0.7, 1 + 0.1 \times 38.11\right) = f(0.7, 4.811)$$ Calculate $f(0.7,4.811)$: $\tan(4.811) = \tan(4.811 - \pi) = \tan(1.67) \approx -11.43$ $$k_3 = \frac{16.444 \times (-11.43) + 9 \times 4.811}{0.7} = \frac{-188.0 + 43.3}{0.7} = \frac{-144.7}{0.7} = -206.7$$ - Calculate $k_4$: $$k_4 = f(0.8, 1 + 0.2 \times (-206.7)) = f(0.8, 1 - 41.34) = f(0.8, -40.34)$$ Calculate $f(0.8,-40.34)$: $e^{3.2} \approx 24.53$, $\tan(-40.34)$ is periodic, reduce modulo $\pi$: $-40.34 + 13\pi \approx -40.34 + 40.84 = 0.5$, so $\tan(-40.34) = \tan(0.5) \approx 0.5463$ $$k_4 = \frac{24.53 \times 0.5463 + 9 \times (-40.34)}{0.8} = \frac{13.39 - 363.06}{0.8} = \frac{-349.67}{0.8} = -437.1$$ - Compute $y_1$: $$y_1 = 1 + \frac{0.2}{6}(43.62 + 2 \times 38.11 + 2 \times (-206.7) + (-437.1))$$ $$= 1 + \frac{0.2}{6}(43.62 + 76.22 - 413.4 - 437.1) = 1 + \frac{0.2}{6}(-730.66) = 1 - 24.36 = -23.36$$ 5. **Step 2: From $x_1=0.8$, $y_1=-23.36$ to $x_2=1.0$** - Calculate $k_1$: $$k_1 = f(0.8, -23.36)$$ $\tan(-23.36)$ reduce modulo $\pi$: $-23.36 + 7\pi = -23.36 + 21.99 = -1.37$, $\tan(-1.37) \approx -5.04$ $$k_1 = \frac{24.53 \times (-5.04) + 9 \times (-23.36)}{0.8} = \frac{-123.7 - 210.2}{0.8} = \frac{-333.9}{0.8} = -417.4$$ - Calculate $k_2$: $$k_2 = f(0.9, -23.36 + 0.1 \times (-417.4)) = f(0.9, -65.1)$$ $e^{3.6} \approx 36.6$, reduce $\tan(-65.1)$ modulo $\pi$: $-65.1 + 21\pi = -65.1 + 65.97 = 0.87$, $\tan(0.87) \approx 1.19$ $$k_2 = \frac{36.6 \times 1.19 + 9 \times (-65.1)}{0.9} = \frac{43.55 - 585.9}{0.9} = \frac{-542.35}{0.9} = -602.6$$ - Calculate $k_3$: $$k_3 = f(0.9, -23.36 + 0.1 \times (-602.6)) = f(0.9, -83.62)$$ Reduce $\tan(-83.62)$ modulo $\pi$: $-83.62 + 27\pi = -83.62 + 84.82 = 1.2$, $\tan(1.2) \approx 2.44$ $$k_3 = \frac{36.6 \times 2.44 + 9 \times (-83.62)}{0.9} = \frac{89.3 - 752.6}{0.9} = \frac{-663.3}{0.9} = -737.0$$ - Calculate $k_4$: $$k_4 = f(1.0, -23.36 + 0.2 \times (-737.0)) = f(1.0, -170.76)$$ $e^{4} \approx 54.6$, reduce $\tan(-170.76)$ modulo $\pi$: $-170.76 + 54\pi = -170.76 + 169.65 = -1.11$, $\tan(-1.11) \approx -2.14$ $$k_4 = \frac{54.6 \times (-2.14) + 9 \times (-170.76)}{1.0} = -116.8 - 1536.8 = -1653.6$$ - Compute $y_2$: $$y_2 = -23.36 + \frac{0.2}{6}(-417.4 + 2(-602.6) + 2(-737.0) + (-1653.6))$$ $$= -23.36 + \frac{0.2}{6}(-417.4 - 1205.2 - 1474 - 1653.6) = -23.36 + \frac{0.2}{6}(-4750.2) = -23.36 - 158.34 = -181.7$$ 6. **Final answer:** The RK4 approximation for $y(1)$ is approximately $$\boxed{-181.7}$$ This large negative value indicates the solution grows rapidly in magnitude and oscillates due to the tangent function.