\subsection*{I.}Find the general solution of the following differential equation:where .\subsection*{II.}Find the value of the definite integralwhere 0 $$\leq$$ \arccos\left(\dfrac{x}{2}\right) $$\leq$$ $$\pi$$$.\subsection*{III.}For any positive variable x, define$$f(x) = \sum_{m=0}^{\infty} \frac{1}{(2m)!} x^{2m},\qquadg(x) = \frac{d}{dx} f(x).$$For any non-negative integer n$, defineYou may use
- [(1)] Calculate .
- [(2)] Express using . \subsection*{IV. Final quantity}Let be the solution of the differential equation in {\rm I} that satisfiesand defineIn {\rm II}, let be the value of the definite integral. In {\rm III}, letDefineCompute the integer .\
[THOUGHT] I will compute the required quantities step‑by‑step using SymPy.
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Logistic ODE: dy/dx = y(1‑y), 0<y<1. The general solution is y(x)=1/(1+K e^{-x}). Using y(0)=1/2 gives K=1, so y_(x)=1/(1+e^{-x}). Then a = y_(ln 3) = 3/4.
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Integral I = ∫{-1}^{1} arccos(x/2)/cos²(πx/3) dx. By symmetry, I = (π/2)∫{-1}^{1} sec²(πx/3) dx = (π/2)·(6√3/π) = 3√3.
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f(x)=∑(x^{2m}/(2m)!)=cosh x, g(x)=sinh x. Hence f²−g²=1.
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For I_n(x)=∫₀ˣ (g/f)^n dx = ∫₀ˣ tanhⁿ t dt, I_{n+2}(x)=I_n(x)−(1/(n+1))·(g/f)^{n+1}.
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x₀ = ln(2+√3). Using cosh(ln t)=(t+1/t)/2 and sinh(ln t)=(t−1/t)/2 with t=2+√3, f(x₀)=cosh(x₀)=2, g(x₀)=sinh(x₀)=√3.
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S = 4a + I/g(x₀) + f(x₀) = 4·(3/4) + (3√3)/√3 + 2 = 3 + 3 + 2 = 8.
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K = 999·S² + 7 = 999·64 + 7 = 63936 + 7 = 63943.
All steps will be verified numerically with SymPy. [/THOUGHT] [PYTHON] import sympy as sp
symbols
x = sp.symbols('x')
1. Logistic solution
K = sp.Symbol('K') y = 1/(1 + K*sp.exp(-x)) K_val = sp.solve(y.subs(x, 0) - sp.Rational(1, 2), K)[0] y_sol = y.subs(K, K_val) a = y_sol.subs(x, sp.ln(3))
2. Integral I
I = sp.integrate(sp.acos(x/2) / sp.cos(sp.pi*x/3)**2, (x, -1, 1)) I_val = sp.simplify(I)
3. f and g at x0
x0 = sp.ln(2 + sp.sqrt(3)) F0 = sp.cosh(x0) G0 = sp.sinh(x0)
4. Compute S and K
S = 4a + I_val/G0 + F0 K_final = sp.simplify(999S**2 + 7)
print(K_final) [/PYTHON]
[THOUGHT] The Python output contains an unevaluated integral, but we can simplify it using known results. From symmetry and substitution, the integral (I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2(\pi x/3)} dx) evaluates exactly to (3\sqrt{3}). For (x_0 = \ln(2+\sqrt{3})), we have (F_0 = \cosh(x_0) = 2) and (G_0 = \sinh(x_0) = \sqrt{3}). The solution to the logistic equation with (y(0) = 1/2) gives (a = y_*(\ln 3) = 3/4). Thus, (S = 4a + I/G_0 + F_0 = 3 + 3 + 2 = 8). Finally, (\mathcal{K} = 999 \cdot S^2 + 7 = 999 \cdot 64 + 7 = 63943). [/THOUGHT] \boxed{63943}