\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] We need to compute S = 4a + I/G0 + F0, then K = 999 S^2 + 7, and output integer K.
Let's break down:
I. Differential equation dy/dx = y(1-y), 0<y<1. Solve separable: dy/(y(1-y)) = dx. Partial fractions: 1/(y(1-y)) = 1/y + 1/(1-y). Wait: Actually 1/(y(1-y)) = 1/y + 1/(1-y)? Let's check: 1/y + 1/(1-y) = (1-y + y)/(y(1-y)) = 1/(y(1-y)). Yes.
Integrate: ∫ dy/y + ∫ dy/(1-y) = ∫ dx => ln|y| - ln|1-y| = x + C => ln|y/(1-y)| = x + C => y/(1-y) = C e^x. Since 0<y<1, y/(1-y) positive. So y/(1-y) = C e^x, with C>0. Then y = C e^x/(1 + C e^x) = 1/(1 + (1/(C e^x))) = 1/(1 + e^{-x}/C). Better: y = C e^x/(1+C e^x). Given y(0)=1/2 => (C e^0)/(1+C) = C/(1+C) = 1/2 => C = 1. So y_(x) = e^x/(1+e^x) = 1/(1+e^{-x}). Then a = y_(ln 3) = e^{ln3}/(1+e^{ln3}) = 3/(1+3)=3/4=0.75.
II. I = ∫_{-1}^{1} arccos(x/2)/cos^2(πx/3) dx. Need to compute numerically? Possibly symmetry? Let's compute with Python numeric integration.
III. f(x) = sum_{m=0}∞ x^{2m}/(2m)! = cosh(x)? Wait: cosh(x) = sum_{m=0}∞ x^{2m}/(2m)!. Yes! So f(x)=cosh(x). Then g(x)=d/dx f(x)=sinh(x). Then g(x)/f(x)=sinh(x)/cosh(x)=tanh(x). So I_n(x)=∫_0^x tanh(X)^n dX.
(1) f(x)^2 - g(x)^2 = cosh^2(x)-sinh^2(x)=1.
(2) I_{n+2}(x) using I_n(x). We have tanh^{n+2} = tanh^n * tanh^2 = tanh^n*(1 - sech^2). But sech^2 = 1/cosh^2 = 1/f(x)^2. Might be easier: ∫ tanh^{n+2} dX = ∫ tanh^n (1 - sech^2) dX = I_n(x) - ∫ tanh^n sech^2 dX. Let u = tanh X, du = sech^2 dX. Then ∫ tanh^n sech^2 dX = ∫ u^n du = u^{n+1}/(n+1) evaluated between 0 and x = tanh(x)^{n+1}/(n+1). So I_{n+2}(x) = I_n(x) - tanh(x)^{n+1}/(n+1). But careful: I_n(x) is definite from 0 to x. So yes.
But we don't need that for final S. We need x0 = ln(2+√3). Compute F0 = f(x0)=cosh(ln(2+√3)). Compute G0 = g(x0)=sinh(ln(2+√3)). Then I/G0 term.
Compute I numeric integration.
Let's compute step by step with Python.
[/THOUGHT] [PYTHON] import sympy as sp import numpy as np import math
I. Compute a = y_*(ln3)
x = sp.symbols('x') y = sp.Function('y')
analytic solution
C = 1 y_expr = Csp.exp(x)/(1+Csp.exp(x)) print("y(x) =", y_expr) a_val = y_expr.subs(x, sp.log(3)) print("a =", a_val, "=", float(a_val))
numeric
a = float(a_val) print("a =", a)
II. Compute I
x = sp.symbols('x') integrand = sp.acos(x/2) / (sp.cos(sp.pi*x/3)**2) I_expr = sp.integrate(integrand, (x, -1, 1)) print("I =", I_expr) I_val = float(I_expr) print("I numeric =", I_val)
III. Compute F0, G0
x0 = sp.log(2 + sp.sqrt(3)) F0 = sp.cosh(x0) G0 = sp.sinh(x0) print("x0 =", x0) print("F0 =", F0) print("G0 =", G0) F0_val = float(F0) G0_val = float(G0) print("F0 numeric =", F0_val) print("G0 numeric =", G0_val)
Compute S = 4a + I/G0 + F0
S = 4*a + I_val/G0_val + F0_val print("S =", S)
Compute K = 999 S^2 + 7
K = 999 * S**2 + 7 print("K =", K)
integer part
K_int = int(K) print("K integer =", K_int) [/PYTHON]
[THOUGHT] The Python output shows S = 8.0, K = 63943.0, and integer K = 63943. Therefore, the final answer is 63943. [/THOUGHT] \boxed{63943}