We are asked to solve a multi-part problem and ultimately compute $\mathcal{K}=999 S^2+7$, where $S=4a+\frac{I}{G_0}+F_0$, with $a$, $I$, $F_0$, $G_ Navier?$ Actually $F_0=f(x_0)$, $G_0=g(x_0)$. We'll go step by step.

I. Differential equation and general solution

Solve $\frac{dy}{dx}=y(1-y)$, $0<y<1$.

Separate variables: $\frac{dy}{y(1-y)}=dx$.
Partial fractions: $\frac{1}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}$.
Integrate: $\ln\left|\frac{y}{1-y}\right|=x+C$ $\implies$ $\frac{y}{1-y}=Ae^{x}$ ($A>0$).
Thus $y=\frac{Ae^{x}}{1+Ae^{x}}=\frac{1}{1+Be^{-x}}$ (with $B=1/A$).
So the general solution is $\boxed{y=\frac{1}{1+Ce^{-x}}}$ (or equivalently $\frac{y}{1-y}=Ce^{x}$).

II. Definite integral $I$

[ I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2\bigl(\frac{\pi}{3}x\bigr)},dx. ]

Let $u=\arccos(x/2)$, $dv=\sec^2(\pi x/3)\,dx$. Then
$du = -\frac{1}{2\sqrt{1-(x/2)^2}}\,dx$, $\; v = \frac{3}{\pi}\tan(\pi x/3)$.

Integration by parts: [ I = \left[ u v \right]{-1}^{1} - \int{-1}^{1} v,du = \Big[ u v \Big|{-1}^{1} \Big] + \frac{3}{2\pi} \int{-1}^{1} \frac{\tan(\pi x/3)}{\sqrt{1-(x/2)^2}},dx. ]

Compute boundary term:

• At $x=1$: $u(1)=\arccos(1/2)=\pi/3$, $v(1)=\frac{3}{\pi}\tan(\pi/3)=\frac{3\sqrt3}{\pi}$ $\Rightarrow$ $u v = \frac{\pi}{3}\cdot\frac{3\sqrt3}{\pi}= \sqrt3$.
• At $x=-1$: $u(-1)=\arccos(-1/2)=2\pi/3$, $v(-1)=\frac{3}{\pi}\tan(-\pi/3)=-\frac{3\sqrt3}{\pi}$ $\Rightarrow$ $u v = \frac{2\pi}{3}\cdot\Big(-\frac{3\sqrt3}{\pi}\Big) = -2\sqrt3$. Thus $[u v]_{-1}^{1} = \sqrt3 - (-2\sqrt3) = 3\sqrt3$.

The remaining integral is odd (since $\tan(\pi x/3)$ is odd and $\sqrt{1-(x/2)^2}$ is even) over $[-1,1]$, so it vanishes. Hence [ I = 3\sqrt3. ]

III. Functions $f$, $g$ and related integrals

Given [ f(x)=\sum_{m=0}^{\infty}\frac{1}{(2m)!}x^{2m}=\cosh x,\qquad g(x)=f'(x)=\sinh x. ]

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

(2) Define $I_n(x)=\int_0^x \left(\frac{g(X)}{f(X)}\right)^n dX = \int_0^x \tanh^n X\,dX$.
Using $\tanh^n = \tanh^{n-2} - \tanh^{n-2}\operatorname{sech}^2$, we obtain [ I_n(x) = I_{n-2}(x) - \frac{1}{n-1}\tanh^{n-1}(x). ] Shifting indices: $I_{n+2}(x) = I_n(x) - \frac{1}{n+1}\left(\frac{g(x)}{f(x)}\right)^{\!n+1}$.

IV. Final computation

• $y_*(x)$ solves $\frac{dy}{dx}=y(1-y)$ with $y(0)=\frac12$. From part I, $\frac{y}{1-y}=e^x$ $\Rightarrow$ $y_*(x)=\frac{1}{1+e^{-x}}$. Then [ a = y_*(\ln 3) = \frac{1}{1+e^{-\ln3}} = \frac{1}{1+\frac13} = \frac34. ]

• For $x_0 = \ln(2+\sqrt3)$, we compute [ f(x_0)=\cosh(\ln(2+\sqrt3)) = \frac{(2+\sqrt3)+(2-\sqrt3)}{2}=2,\qquad g(x_0)=\sinh(\ln(2+\sqrt3)) = \frac{(2+\sqrt3)-(2-\sqrt3)}{2}= \sqrt3. ] Thus $F_0 = 2$, $G_0 = \sqrt3$.

Now [ I = 3\sqrt3,\quad \frac{I}{G_0} = \frac{3\sqrt3}{\sqrt3}=3. ]

Hence [ S = 4a + \frac{I}{G_0} + F_0 = 4\cdot\frac34 + 3 + 2 = 3 + 3 + 2 = 8. ]

Finally, [ \mathcal{K} = 999 S^2 + 7 = 999\cdot 64 + 7 = 63936 + 7 = 63943. ]

[ \boxed{63943} ]

[THOUGHT] We need to compute the integer $\mathcal{K} = 999S^2 + 7$ where $S = 4a + \frac{I}{G_0} + F_0$.

• $a = y_*(\ln 3)$ from the differential equation $\frac{dy}{dx}=y(1-y)$ with $y(0)=\frac12$.
  The solution is $y_*(x) = \frac{1}{1+e^{-x}}$, so $a = \frac{1}{1+e^{-\ln 3}} = \frac{1}{1+\frac13} = \frac34$.
• $I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2\bigl(\frac{\pi}{3}x\bigr)}\,dx$.
  Using integration by parts one finds $I = 3\sqrt{3}$ (verified numerically below).
• $x_0 = \ln(2+\sqrt{3})$, $F_0 = f(x_0) = \cosh x_0$, $G_0 = g(x_0) = \sinh x_0$.
  Exact values: $\cosh(\ln(2+\sqrt{3})) = 2$, $\sinh(\ln(2+\sqrt{3})) = \sqrt{3}$.

Thus $S = 4\cdot\frac34 + \frac{3\sqrt{3}}{\sqrt{3}} + 2 = 3 + 3 + 2 = 8$, and $\mathcal{K} = 999\cdot 8^2 + 7 = 63943$.

To verify the intermediate values numerically, we use Python.

[PYTHON] import math from scipy.integrate import quad

Part I: a

a = 3/4

Part II: I

def integrand(x): return math.acos(x/2) / (math.cos(math.pi * x / 3) ** 2)

I_val, err = quad(integrand, -1, 1) print(f"I = {I_val}")

Part III: x0, F0, G0

x0 = math.log(2 + math.sqrt(3)) F0 = math.cosh(x0) G0 = math.sinh(x0) print(f"F0 = {F0}, G0 = {G0}")

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

K = 999 * S**2 + 7 print(f"K = {K}")

Exact checks

print(f"Exact S = 8") print(f"Exact K = {999*64 + 7}") [/PYTHON]

\boxed{63943}

[THOUGHT] The problem involves multiple steps, primarily focused on computing a final value of $\mathcal{K} = 999S^2 + 7$, where S = 4a + \. We need to compute:

• $a = y_*(\ln 3)$, with $y_*(x)$ being the specific solution to $\frac{dy}{dx} = y(1-y)$ satisfying $y(0) = \frac{1}{2}$.
• $I = \int_{-1}^{1} \frac{\arccos(x/2)}{\cos^2(\pi x/3)}\,dx$,
• $F_0 = f(x_0)$, $G_0 = g(x_0)$, where $f(x) = \sum_{m=0}^\infty \frac{x^{2m}}{(2m)!}$, $g(x) = f'(x)$, and $x_0 = \ln(2 + \sqrt{3})$.

From earlier analysis:

• The general solution to $\frac{dy}{dx} = y(1-y)$ is $y = \frac{1}{1 + Ce^{-x}}$. Using $y(0) = \frac{1}{2}$, we find $C = 1$, so $y_*(x) = \frac{1}{1 + e^{-x}}$, hence $a = y_*(\ln 3) = \frac{3}{4}$.
• The integral $I$ was evaluated via integration by parts and symmetry, yielding $I = 3\sqrt{3}$.
• $f(x) = \cosh x$, $g(x) = \sinh x$, so $F_0 = \cosh(x_0)$, $G_0 = \sinh(x_0)$. With $x_0 = \ln(2 + \sqrt{3})$, we find $F_0 = 2$, $G_0 = \sqrt{3}$.

Thus:

Finally, $\mathcal{K} = 999 \cdot 8^2 + 7 = 63943$.

To verify, we use Python to numerically compute the integral and confirm the values of $F_0$, $G_0$, and $S$.

[PYTHON] import math from scipy.integrate import quad

Compute integral I

def integrand(x): return math.acos(x / 2) / (math.cos(math.pi * x / 3) ** 2)

I_val, err = quad(integrand, -1, 1) print(f"I = {I_val}")

Compute F0 and G0

x0 = math.log(2 + math.sqrt(3)) F0 = math.cosh(x0) G0 = math.sinh(x0) print(f"F0 = {F0}, G0 = {G0}")

Compute S

a = 3/4 S = 4*a + I_val/G0 + F0 print(f"S = {S}")

Compute K

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

[OUTCOME] I = 5.196152422706632 F0 = 1.9999999999999998, G0 = 1.732050807568877 S = 8.0 K = 63943.0 [REASONING] The Python output confirms:

• $I \approx 3\sqrt{3} \approx 5.19615$,
• $F_0 \approx 2$, $G_0 \approx \sqrt{3} \approx 1.73205$,
• $S = 8.0$, so $\mathcal{K} = 999 \cdot 64 + 7 = 63943$.

Thus, the integer $\mathcal{K}$ is exactly 63943. </think>

63943

I've computed the values from the Python output:

• ( I = 3\sqrt{3} \approx 5.196152422706632 )
• ( F_0 = 2 )
• ( G_0 = \sqrt{3} \approx 1.732050807568877 )
• ( S = 4a + \frac{I}{G_0} + F_0 = 8 )
• ( \mathcal{K} = 999 \times 8^2 + 7 = 63943 )

Thus the final answer is:

\boxed{63943}