\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 solve part I of the problem by finding the general solution to the differential equation (\frac{dy}{dx} = y(1-y)) for (0 < y < 1). This is a separable equation. I will use the sympy library to derive and output the solution symbolically. [/THOUGHT] [PYTHON] from sympy import symbols, exp x, C = symbols('x C') y = exp(x) / (exp(x) + C) print(y) [/PYTHON]
[THOUGHT] After solving the differential equation in part I, I found the specific solution ( y_(x) = \frac{e^x}{1 + e^x} ). At ( x = \ln 3 ), ( a = y_(\ln 3) = \frac{3}{4} ).
For part II, I computed the integral ( I = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} dx = 3\sqrt{3} ).
In part III, I calculated:
- Part (1): ( f(x)^2 - g(x)^2 = 1 )
- Part (2): ( I_{n+2}(x) = I_n(x) - \frac{\tanh^{n+1}(x)}{n+1} )
Given ( x_0 = \ln(2+\sqrt{3}) ), ( F_0 = f(x_0) = 2 ), and ( G_0 = g(x_0) = \sqrt{3} ), I computed:
- ( \frac{I}{G_0} = \frac{3\sqrt{3}}{\sqrt{3}} = 3 )
- ( S = 4a + \frac{I}{G_0} + F_0 = 4 \cdot \frac{3}{4} + 3 + 2 = 3 + 3 + 2 = 8 )
- ( \mathcal{K} = 999S^2 + 7 = 999 \cdot 64 + 7 = 63840 + 7 = 63847 )
The integer ( \mathcal{K} ) is 63847. [/THOUGHT] \boxed{63847}