[THOUGHT] I need to solve this multi-part problem step by step. Let's start with part I, which is solving the differential equation dy/dx = y(1-y). This is a separable differential equation. I'll solve it using separation of variables and then verify the solution using Python. [/THOUGHT] [PYTHON] from sympy import symbols, Function, dsolve, Eq, log x = symbols('x') y = Function('y')(x) ode = Eq(y.diff(x), y*(1 - y)) solution = dsolve(ode, y) print(solution) [/PYTHON]

[THOUGHT] Now, for part II, I need to evaluate the definite integral I = ∫_{-1}^{1} arccos(x/2)/cos²(πx/3) dx. This looks complex, but perhaps substitution or symmetry can help. I'll attempt to compute it numerically first to get an idea. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, acos, cos, pi x = symbols('x') integrand = acos(x/2) / (cos(pi*x/3)**2) integral = integrate(integrand, (x, -1, 1)) print(integral) [/PYTHON]

[THOUGHT] For part III, I need to work with the functions f(x) and g(x). First, I recognize that f(x) is the Taylor series expansion of the hyperbolic cosine function, cosh(x). Then g(x) is its derivative, which is sinh(x). I'll verify this and then proceed to compute f(x)^2 - g(x)^2 and find the recurrence relation for I_{n+2}(x). [/THOUGHT] [PYTHON] from sympy import symbols, sinh, cosh, integrate, exp x = symbols('x') f = cosh(x) g = sinh(x) part1 = f2 - g2 print(part1) [/PYTHON]

[THOUGHT] Now, I need to find the recurrence relation for I_{n+2}(x). Given that f(x)^2 - g(x)^2 = 1, I can use this identity to relate I_{n+2}(x) to I_n(x). I'll derive the relation and then verify it. [/THOUGHT] [PYTHON]

Since f^2 - g^2 = 1, we can write (g/f)^2 = 1 - (f/g)^2, but this might not directly help.

Alternatively, note that d/dx (g/f) = (fg' - gf') / f^2 = (f^2 + g^2)/f^2 = 1 + (g/f)^2

But I need to find a recurrence for I_{n+2}(x) = ∫ (g/f)^{n+2} dx.

Let me think differently: I_{n+2}(x) = ∫ (g/f)^n * (g/f)^2 dx = ∫ (g/f)^n * (1 - f^2/g^2) dx, but this seems circular.

Perhaps integration by parts or substitution is needed. For now, I'll proceed to part IV.

[/THOUGHT]

[THOUGHT] For part IV, I need to compute the final quantity K. First, I'll find y_(x) from part I with y_(0) = 1/2. Then compute a = y_*(ln 3). Next, use the integral I from part II and the values F_0 and G_0 from part III at x_0 = ln(2 + sqrt(3)). Finally, compute S and K. [/THOUGHT] [PYTHON] from sympy import log, exp, sqrt, N

Part I: Solve y(0) = 1/2

C1 = symbols('C1') general_solution = solution.rhs particular_solution = general_solution.subs(C1, log(1/2 / (1 - 1/2))) a = particular_solution.subs(x, log(3))

Part II: Integral I

I_value = integral

Part III: x_0 = ln(2 + sqrt(3))

x0 = log(2 + sqrt(3)) F0 = cosh(x0) G0 = sinh(x0)

Compute S

S = 4*a + I_value/G0 + F0 K = 999 * S**2 + 7 K_evaluated = N(K) print(K_evaluated) [/PYTHON]

[THOUGHT] The Python code for part IV had a syntax error. I need to fix it by properly defining the symbols and expressions. Let me correct the code and rerun it to compute the final quantity K. [/THOUGHT] [PYTHON] from sympy import symbols, log, exp, sqrt, N, Integral, acos, cos, pi, cosh, sinh x = symbols('x') C1 = symbols('C1')

Part I: Solve y(0) = 1/2

general_solution = 1 / (C1*exp(-x) + 1) particular_solution = general_solution.subs(C1, log(1/2 / (1 - 1/2))) a = particular_solution.subs(x, log(3))

Part II: Integral I

integrand = acos(x/2) / (cos(pi*x/3)**2) I_value = Integral(integrand, (x, -1, 1)).doit()

Part III: x_0 = ln(2 + sqrt(3))

x0 = log(2 + sqrt(3)) F0 = cosh(x0) G0 = sinh(x0)

Compute S

S = 4*a + I_value/G0 + F0 K = 999 * S**2 + 7 K_evaluated = N(K) print(K_evaluated) [/PYTHON]

[THOUGHT] The final computed value of (\mathcal{K}) is 80926.0000000000, which is an integer. Thus, the answer is: [/THOUGHT] \boxed{80926}