[THOUGHT] Let's solve each part of the problem step-by-step using Python code verification.

Part I: Differential Equation

The given differential equation is: [ \frac{dy}{dx} = y(1-y) ] This is a separable differential equation. We can rewrite it as: [ \frac{1}{y(1-y)} , dy = dx ] Using partial fractions, we get: [ \frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y} ] So the equation becomes: [ \left( \frac{1}{y} + \frac{1}{1-y} \right) dy = dx ] Integrating both sides: [ \int \left( \frac{1}{y} + \frac{1}{1-y} \right) dy = \int dx ] [ \ln |y| - \ln |1-y| = x + C ] [ \ln \left| \frac{y}{1-y} \right| = x + C ] Exponentiating both sides: [ \left| \frac{y}{1-y} \right| = e^{x+C} ] Let ( e^C = K ), then: [ \frac{y}{1-y} = K e^x ] Solving for ( y ): [ y = \frac{K e^x}{1 + K e^x} ] Given ( 0 < y < 1 ), we can drop the absolute value.

Part II: Definite Integral

The given integral is: [ I = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} , dx ] To evaluate this integral, we will use numerical integration in Python.

Part III: Series Functions

(1) Calculate ( f(x)^2 - g(x)^2 )

Given: [ f(x) = \sum_{m=0}^{\infty} \frac{1}{(2m)!} x^{2m},\qquad g(x) = \frac{d}{dx} f(x) ] First, let's find ( g(x) ): [ g(x) = \sum_{m=1}^{\infty} \frac{2m}{(2m)!} x^{2m-1} = \sum_{m=1}^{\infty} \frac{1}{(2m-1)!} x^{2m-1} ] We recognize that: [ f(x) = \cosh(x),\qquad g(x) = \sinh(x) ] Using the hyperbolic identity: [ \cosh^2(x) - \sinh^2(x) = 1 ] So: [ f(x)^2 - g(x)^2 = 1 ]

(2) Express ( I_{n+2}(x) ) using ( I_n(x) )

Given: [ I_n(x) = \int_0^x \left(\frac{g(X)}{f(X)}\right)^n , dX ] We know: [ \frac{g(x)}{f(x)} = \frac{\sinh(x)}{\cosh(x)} = \tanh(x) ] So: [ I_n(x) = \int_0^x \tanh^n(X) , dX ]

Using integration by parts for ( I_{n+2}(x) ): Let ( u = \tanh^{n+1}(X) ) and ( dv = \tanh(X) , dX ). Then ( du = (n+1) \tanh^n(X) \text{sech}^2(X) , dX ) and ( v = \ln(\cosh(X)) ).

Applying integration by parts: [ I_{n+2}(x) = \int_0^x \tanh^{n+2}(X) , dX = \left[ \tanh^{n+1}(X) \ln(\cosh(X)) \right]0^x - \int_0^x (n+1) \tanh^n(X) \text{sech}^2(X) \ln(\cosh(X)) , dX ] Since ( \tanh(0) = 0 ) and ( \ln(\cosh(0)) = \ln(1) = 0 ): [ I{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \int_0^x \tanh^n(X) \text{sech}^2(X) \ln(\cosh(X)) , dX ] Notice that: [ \text{sech}^2(X) = 1 - \tanh^2(X) ] Thus: [ I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \left( \int_0^x \tanh^n(X) \ln(\cosh(X)) , dX - \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) , dX \right) ] Let ( J_n(x) = \int_0^x \tanh^n(X) \ln(\cosh(X)) , dX ): [ I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) J_n(x) + (n+1) J_{n+2}(x) ] Since ( J_{n+2}(x) ) contains ( I_{n+2}(x) ) and ( J_n(x) ) contains ( I_n(x) ), we focus on simplifying: [ I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) J_n(x) + (n+1) \left( \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) , dX \right) ] [ I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \left( \int_0^x \tanh^n(X) \ln(\cosh(X)) , dX - \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) , dX \right) ] [ I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) I_n(x) + n I_{n+2}(x) ] [ I_{n+2}(x) = \frac{1}{n+1} \tanh^{n+1}(x) \ln(\cosh(x)) - I_n(x) ]

Part IV: Final Quantity

First, find ( y_(x) ) that satisfies ( y_(0) = \frac{1}{2} ): [ y_(x) = \frac{K e^x}{1 + K e^x} ] Using the initial condition: [ \frac{1}{2} = \frac{K}{1 + K} ] [ 1 + K = 2K ] [ K = 1 ] Thus: [ y_(x) = \frac{e^x}{1 + e^x} ] Now find ( a = y_(\ln 3) ): [ a = y_(\ln 3) = \frac{e^{\ln 3}}{1 + e^{\ln 3}} = \frac{3}{1 + 3} = \frac{3}{4} ]

Next, evaluate the integral ( I ) using Python: [ I = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} , dx ]

Then, compute ( F_0 ) and ( G_0 ) where ( x_0 = \ln(2+\sqrt{3}) ): [ F_0 = f(x_0) = \cosh(x_0),\quad G_0 = g(x_0) = \sinh(x_0) ] Finally, compute ( S ) and ( \mathcal{K} ).

Let's execute the Python code for these calculations. [/THOUGHT] [PYTHON] import numpy as np from scipy.integrate import quad

Part I: Differential Equation

We already have the solution y_*(x) = e^x / (1 + e^x)

a = y_*(ln 3)

a = 3 / 4

Part II: Definite Integral

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

I, _ = quad(integrand, -1, 1)

Part III: Series Functions

x_0 = ln(2 + sqrt(3))

x_0 = np.log(2 + np.sqrt(3))

F_0 = f(x_0) = cosh(x_0)

F_0 = np.cosh(x_0)

G_0 = g(x_0) = sinh(x_0)

G_0 = np.sinh(x_0)

Part IV: Final Quantity

S = 4 * a + I / G_0 + F_0 K = 999 * S**2 + 7

print(K)

[THOUGHT] The Python output confirms the final value of (\mathcal{K}) to be 63943. Therefore, the integer (\mathcal{K}) is \boxed{63943}.