\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. The given differential equation is separable: dy/dx = y(1-y) => dy/y(1-y) = dx Integrating both sides, ln|y/(1-y)| = x + C (say) Since 0 < y < 1, |y/(1-y)| = y/(1-y), so we have ln(y/(1-y)) = x + C => y/(1-y) = e^(x+C) = A (say) Here A is a positive constant. Solving for y, y = A/(A+y)
II. The given integral can be solved by substitution: u = arccos(x/2), dv = 1/(cos^2((π/3)x)) dx Then du/dx = -1/(2cos((π/3)x)sin((π/3)x)), v = (-(π/6)sin((π/3)x))/(cos((π/3)x))^3 Integrating by parts, ∫(arccos(x/2))/(cos^2((π/3)x)) dx = [-(Aπ/6)sin((π/3)x)cos((π/3)x)^3]/(-1/4) | from -1 to 1 = (Aπ/12)(1 - (-1/2)) = Aπ/6
Here, u=-1 when x=-1 and u=0 when x=1. So A = π/2. Hence I = (∏/6)/2 = π/12.
III. (1) f(x)^2 - g(x)^2 = (Σ(1/(2m)!)x^(2m))^2 - (Σ((2m)(1/(2m)!)x^(2m-1)))^2 = Σ(1/(2m)!(4m^2)x^(4m-2)) - Σ(1/(2m-1)!(2m)^2x^(4m-2)) = Σ(1/(2m)!*(4m^2 - (2m)^2)*x^(4m-2)) = Σ(1/(2m)!2m(2m-1)*x^(4m-2))
(2) I_(n+2)(x) = ∫_0^x [(g(X))/(f(X))]^(n+2) dX = ∫_0^x [Σ((2(n+2)m/(2m)!)X^(2m-1))] / [Σ(1/(2m)!X^(2m))]^(n+2) dX = ∫_0^x Σ[(2(n+2)m/(2m)!)(1/Σ(1/(2m)!)X^(2m-n))] dX = Σ[(2(n+2)m/(2m)!)∫_0^x (1/X^(n)) dX] = Σ[(2(n+2)m/(2m)!)(-1/n)[1 - (1/(1+x^n))] from 0 to x] = Σ[-(2(n+2)mn)/(n!(2m-n))[1 - (1/(1+x^n))] = Σ[-(2(n+2)mn)/(n!(2m-n))] - Σ[(2(n+2)mn)/(n!(1+x^n)(2m-n))] = I_n(x) - Σ[(2(n+2)mn)/(n!(1+x^n)(2m-n!))]
IV. I. The differential equation's solution y*(x) with initial condition y_(0)=1/2 is y = 1/(A+y), where A=1 (from ln(1/(1-y))=0). Hence, y_(ln3) = 1/(1+3^(1/2)) = sqrt(3)/2.
II. From II above, I = π/12.
III. f(x_0) = Σ(1/(2m)!)(2x_0)^(2m) g(x_0) = Σ((2m)/(2m)!)(2x_0)^(2m-1)
G_0 = g(x_0) with x_0=ln(2+sqrt(3)) = 2Σ(m/(2m)!)[(2+sqrt(3))^m] = 2Σ(m/(2m)!)(2^m)*(1+(sqrt(3)/4))^(m) ≈ 5.83 (from first few terms)
F_0 = f(x_0) = Σ(1/(2m)!)*(2ln(2+sqrt(3)))^(2m) ≈ 11.55
S = 4*sqrt(3)/2 + π/12/G_0 + F_0 ≈ 15.77
K = 999*S^2 + 7 = 1520986
\boxed{1520986}