[THOUGHT] Let me solve this problem step by step.

Part I: Differential Equation The equation is $\frac{dy}{dx} = y(1-y)$, which is separable. $\frac{dy}{y(1-y)} = dx$

Using partial fractions: $\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}$

Integrating: $\ln|y| - \ln|1-y| = x + C$ $\ln\frac{y}{1-y} = x + C$ $\frac{y}{1-y} = Ae^x$ where $A = e^C$ $y = Ae^x(1-y)$ $y = Ae^x - Ae^xy$ $y(1 + Ae^x) = Ae^x$ $y = \frac{Ae^x}{1 + Ae^x}$

With $y_*(0) = \frac{1}{2}$: $\frac{A}{1+A} = \frac{1}{2}$, so $A = 1$. Thus $y_*(x) = \frac{e^x}{1 + e^x}$

$a = y_*(\ln 3) = \frac{3}{1+3} = \frac{3}{4}$

Part II: Definite Integral $I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2(\frac{\pi}{3}x)} dx$

Part III: Series $f(x) = \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!} = \cosh(x)$ $g(x) = f'(x) = \sinh(x)$

(1) $f(x)^2 - g(x)^2 = \cosh^2(x) - \sinh^2(x) = 1$

(2) For $I_n(x) = \int_0^x \tanh^n(X) dX$

$I_{n+2}(x) = \int_0^x \tanh^{n+2}(X) dX = \int_0^x \tanh^n(X) \cdot \tanh^2(X) dX$ $= \int_0^x \tanh^n(X)(1 - \text{sech}^2(X)) dX$ $= I_n(x) - \int_0^x \tanh^n(X) \text{sech}^2(X) dX$ $= I_n(x) - \frac{\tanh^{n+1}(x)}{n+1}$

For $x_0 = \ln(2+\sqrt{3})$: $e^{x_0} = 2 + \sqrt{3}$ $F_0 = \cosh(x_0) = \frac{e^{x_0} + e^{-x_0}}{2}$ $G_0 = \sinh(x_0) = \frac{e^{x_0} - e^{-x_0}}{2}$

Let me compute these numerically. [/THOUGHT] [PYTHON] import numpy as np from scipy import integrate

Part I: a = y_*(ln 3)

a = 3 / (1 + 3) print(f"a = {a}")

Part II: Definite integral

def integrand(x): return np.arccos(x/2) / (np.cos(np.pi * x / 3))**2

I, error = integrate.quad(integrand, -1, 1) print(f"I = {I}")

Part III: x_0, F_0, G_0

x0 = np.log(2 + np.sqrt(3)) print(f"x0 = {x0}")

F0 = np.cosh(x0) G0 = np.sinh(x0) print(f"F0 = {F0}") print(f"G0 = {G0}")

Verify with explicit formula

e_x0 = 2 + np.sqrt(3) e_neg_x0 = 1 / e_x0 F0_check = (e_x0 + e_neg_x0) / 2 G0_check = (e_x0 - e_neg_x0) / 2 print(f"F0_check = {F0_check}") print(f"G0_check = {G0_check}")

Calculate S

S = 4*a + I/G0 + F0 print(f"S = {S}")

Calculate K

K = 999 * S**2 + 7 print(f"K = {K}") print(f"K rounded = {round(K)}") [/PYTHON]

[THOUGHT] The Python calculation confirms all the values:

• $a = y_*(\ln 3) = \frac{3}{4} = 0.75$
• $I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2(\frac{\pi}{3}x)} dx \approx 5.196152... = 3\sqrt{3}$
• $x_0 = \ln(2+\sqrt{3})$
• $F_0 = \cosh(x_0) = 2$
• $G_0 = \sinh(x_0) = \sqrt{3}$

Therefore: $S = 4a + \frac{I}{G_0} + F_0 = 4 \cdot \frac{3}{4} + \frac{3\sqrt{3}}{\sqrt{3}} + 2 = 3 + 3 + 2 = 8$

$\mathcal{K} = 999 \cdot S^2 + 7 = 999 \cdot 64 + 7 = 63936 + 7 = 63943$ [/THOUGHT]

$\boxed{63943}$