Problem Analysis #6

Solar-Pro-2 (31B)(high) · Run 1

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Problem Statement

\subsection*{I.}Find the general solution y(x)y(x) of the following differential equation:

dydx=y(1−y),\frac{dy}{dx} = y(1-y),

where 0<y<10 < y < 1.\subsection*{II.}Find the value of the definite integral

I=∫−11arccos⁡(x2)cos⁡2(π3x) dx,I = \int_{-1}^{1} \frac{\arccos\left(\dfrac{x}{2}\right)}{\cos^2\left(\dfrac{\pi}{3}x\right)} \, dx,

where 0≤arccos⁡(x2)≤π0 \leq \arccos\left(\dfrac{x}{2}\right) \leq \pi.\subsection*{III.}For any positive variable xx, define

f(x)=∑m=0∞1(2m)!x2m,\qquadg(x)=ddxf(x).f(x) = \sum_{m=0}^{\infty} \frac{1}{(2m)!} x^{2m},\qquadg(x) = \frac{d}{dx} f(x).

For any non-negative integer nn, define

In(x)=∫0x(g(X)f(X))n dX.I_n(x) = \int_0^x \left(\frac{g(X)}{f(X)}\right)^n \, dX.

You may use

exp⁡(x)=∑m=0∞xmm!.\exp(x) = \sum_{m=0}^{\infty} \frac{x^m}{m!}.
  • [(1)] Calculate f(x)2−g(x)2f(x)^2 - g(x)^2.
  • [(2)] Express In+2(x)I_{n+2}(x) using In(x)I_n(x). \subsection*{IV. Final quantity}Let y∗(x)y_*(x) be the solution of the differential equation in {\rm I} that satisfies
y∗(0)=12,y_*(0) = \frac{1}{2},

and define

a:=y∗(ln⁡3).a := y_*(\ln 3).

In {\rm II}, let II be the value of the definite integral. In {\rm III}, let

x0=ln⁡(2+3),\qquadF0:=f(x0),G0:=g(x0).x_0 = \ln(2+\sqrt{3}),\qquadF_0 := f(x_0),\quad G_0 := g(x_0).

Define

S:=4a+IG0+F0,K:=999 S2+7.S := 4a + \frac{I}{G_0} + F_0,\qquad\mathcal{K} := 999\,S^2 + 7.

Compute the integer K\mathcal{K}.\

Ground Truth
63943
Execution Trace
Reasoning Process

[THOUGHT] To solve the given problem, we will address each part step-by-step and use Python to verify critical calculations.

Part I: Solving the Differential Equation The differential equation dydx=y(1−y)\frac{dy}{dx} = y(1 - y) is separable. Separating variables and integrating gives:

∫1y(1−y) dy=∫1 dx\int \frac{1}{y(1 - y)} \, dy = \int 1 \, dx

Using partial fractions, we decompose the left-hand side:

1y(1−y)=1y+11−y\frac{1}{y(1 - y)} = \frac{1}{y} + \frac{1}{1 - y}

Integrating both sides:

ln⁡∣y∣−ln⁡∣1−y∣=x+C  ⟹  ln⁡∣y1−y∣=x+C\ln|y| - \ln|1 - y| = x + C \implies \ln\left|\frac{y}{1 - y}\right| = x + C

Exponentiating both sides:

y1−y=Cex  ⟹  y=Cex1+Cex\frac{y}{1 - y} = Ce^x \implies y = \frac{Ce^x}{1 + Ce^x}

Applying the initial condition y(0)=12y(0) = \frac{1}{2}:

12=C1+C  ⟹  C=1\frac{1}{2} = \frac{C}{1 + C} \implies C = 1

Thus, the solution is:

y∗(x)=ex1+exy_*(x) = \frac{e^x}{1 + e^x}

For a=y∗(ln⁡3)a = y_*(\ln 3):

a=eln⁡31+eln⁡3=31+3=34a = \frac{e^{\ln 3}}{1 + e^{\ln 3}} = \frac{3}{1 + 3} = \frac{3}{4}

Part II: Evaluating the Definite Integral The integral is:

I=∫−11arccos⁡(x2)cos⁡2(π3x) dxI = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx

Using the symmetry of the integrand and the identity arccos⁡(−z)=π−arccos⁡(z)\arccos(-z) = \pi - \arccos(z), we split the integral:

I=∫−10arccos⁡(x2)cos⁡2(π3x) dx+∫01arccos⁡(x2)cos⁡2(π3x) dxI = \int_{-1}^{0} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx + \int_{0}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx

Substituting x=−tx = -t in the first integral:

∫−10arccos⁡(x2)cos⁡2(π3x) dx=∫10π−arccos⁡(t2)cos⁡2(π3t)(−dt)=∫01π−arccos⁡(t2)cos⁡2(π3t) dt\int_{-1}^{0} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx = \int_{1}^{0} \frac{\pi - \arccos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{\pi}{3}t\right)} (-dt) = \int_{0}^{1} \frac{\pi - \arccos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{\pi}{3}t\right)} \, dt

Combining the integrals:

I=∫01π−arccos⁡(t2)cos⁡2(π3t) dt+∫01arccos⁡(t2)cos⁡2(π3t) dt=π∫011cos⁡2(π3t) dtI = \int_{0}^{1} \frac{\pi - \arccos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{\pi}{3}t\right)} \, dt + \int_{0}^{1} \frac{\arccos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{\pi}{3}t\right)} \, dt = \pi \int_{0}^{1} \frac{1}{\cos^2\left(\frac{\pi}{3}t\right)} \, dt

The integral of sec⁡2(u)\sec^2(u) is tan⁡(u)\tan(u):

I=π[3πtan⁡(π3t)]01=3(tan⁡(π3)−tan⁡(0))=33I = \pi \left[ \frac{3}{\pi} \tan\left(\frac{\pi}{3}t\right) \right]_{0}^{1} = 3 \left( \tan\left(\frac{\pi}{3}\right) - \tan(0) \right) = 3 \sqrt{3}

Part III: Hyperbolic Functions and Integrals Given:

f(x)=∑m=0∞x2m(2m)!=cosh⁡(x),g(x)=ddxf(x)=sinh⁡(x)f(x) = \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!} = \cosh(x), \quad g(x) = \frac{d}{dx} f(x) = \sinh(x)
  1. Calculate f(x)2−g(x)2f(x)^2 - g(x)^2:
f(x)2−g(x)2=cosh⁡2(x)−sinh⁡2(x)=1f(x)^2 - g(x)^2 = \cosh^2(x) - \sinh^2(x) = 1
  1. Express In+2(x)I_{n+2}(x) using In(x)I_n(x):
In(x)=∫0x(g(X)f(X))n dX=∫0xtanh⁡n(X) dXI_n(x) = \int_0^x \left(\frac{g(X)}{f(X)}\right)^n \, dX = \int_0^x \tanh^n(X) \, dX

Using the identity tanh⁡2(X)=1−sech2(X)\tanh^2(X) = 1 - \text{sech}^2(X):

In+2(x)=∫0xtanh⁡n(X)(1−sech2(X)) dX=In(x)−∫0xtanh⁡n(X)sech2(X) dXI_{n+2}(x) = \int_0^x \tanh^n(X) (1 - \text{sech}^2(X)) \, dX = I_n(x) - \int_0^x \tanh^n(X) \text{sech}^2(X) \, dX

Let u=tanh⁡(X)u = \tanh(X), then du=sech2(X) dXdu = \text{sech}^2(X) \, dX:

∫0xtanh⁡n(X)sech2(X) dX=∫0tanh⁡(x)un du=tanh⁡n+1(x)n+1\int_0^x \tanh^n(X) \text{sech}^2(X) \, dX = \int_0^{\tanh(x)} u^n \, du = \frac{\tanh^{n+1}(x)}{n+1}

Thus:

In+2(x)=In(x)−tanh⁡n+1(x)n+1I_{n+2}(x) = I_n(x) - \frac{\tanh^{n+1}(x)}{n+1}

Part IV: Final Calculation Given:

x0=ln⁡(2+3),F0=f(x0)=cosh⁡(x0),G0=g(x0)=sinh⁡(x0)x_0 = \ln(2 + \sqrt{3}), \quad F_0 = f(x_0) = \cosh(x_0), \quad G_0 = g(x_0) = \sinh(x_0)

Using the properties of hyperbolic functions:

cosh⁡(ln⁡(2+3))=2,sinh⁡(ln⁡(2+3))=3\cosh(\ln(2 + \sqrt{3})) = 2, \quad \sinh(\ln(2 + \sqrt{3})) = \sqrt{3}

Thus:

F0=2,G0=3F_0 = 2, \quad G_0 = \sqrt{3}

Now, compute SS:

S=4a+IG0+F0=4(34)+333+2=3+3+2=8S = 4a + \frac{I}{G_0} + F_0 = 4 \left(\frac{3}{4}\right) + \frac{3\sqrt{3}}{\sqrt{3}} + 2 = 3 + 3 + 2 = 8

Finally, compute K\mathcal{K}:

K=999S2+7=999×64+7=63936+7=63943\mathcal{K} = 999S^2 + 7 = 999 \times 64 + 7 = 63936 + 7 = 63943

[PYTHON]

Verification of key values

import math

Part I: a = y*(ln 3)

a = math.exp(math.log(3)) / (1 + math.exp(math.log(3))) a = 3 / 4 # Simplified

Part II: I = 3 * sqrt(3)

I = 3 * math.sqrt(3)

Part III: F0 and G0

x0 = math.log(2 + math.sqrt(3)) F0 = math.cosh(x0) G0 = math.sinh(x0)

Part IV: Compute S and K

S = 4 * a + I / G0 + F0 K = 999 * S**2 + 7

print(f"a = {a}") print(f"I = {I}") print(f"F0 = {F0}, G0 = {G0}") print(f"S = {S}") print(f"K = {K}") [/PYTHON]

Tool Output
[PYTHON OUTPUT] a = 0.75 I = 5.196152422706632 F0 = 1.9999999999999998, G0 = 1.732050807568877 S = 8.0 K = 63943.0 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python output confirms the calculations. The value of aa is 0.750.75, II is 33≈5.1963\sqrt{3} \approx 5.196, F0=2F_0 = 2, and G0=3G_0 = \sqrt{3}. Thus, S=4(0.75)+333+2=8S = 4(0.75) + \frac{3\sqrt{3}}{\sqrt{3}} + 2 = 8. Finally, K=999×82+7=63943\mathcal{K} = 999 \times 8^2 + 7 = 63943.

63943\boxed{63943}
Final Model Output
63943

Comparative Analysis (P6) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
330.0
GPT-5.2 (high)
111.5
Kanana-2-30B-Thinking-2601
76.9
Solar-Open-100B
67.2
GPT-5.1 (high)
57.7
HCX-007(high)
55.6
Llama-VARCO-8B-Instruct
54.7
Deepseek-V3.2
48.3
Gemini-3-Pro-Preview
47.6
A.X-4.0 (72B)
45.4
Kanana-2-30B-Thinking
42.7
Solar-Pro-2 (31B)(high)
40.2
Grok-4.1-fast
29.0
K-EXAONE-236B-A23B
22.8
Claude-Opus-4.5
20.6
EXAONE-4.0.1-32B (high)
11.1

Avg Token Usage

K-EXAONE-236B-A23B
39,933
GPT-5.2 (high)
26,193
Kanana-2-30B-Thinking-2601
19,787
Solar-Open-100B
12,619
HCX-007(high)
10,581
EXAONE-4.0.1-32B (high)
9,149
Solar-Pro-2 (31B)(high)
9,027
GPT-5.1 (high)
8,547
Grok-4.1-fast
8,351
Kanana-2-30B-Thinking
8,050
Gemini-3-Pro-Preview
7,889
A.X-4.0 (72B)
7,410
K-EXAONE-236B-A23B
6,804
Claude-Opus-4.5
5,887
Deepseek-V3.2
5,395
Llama-VARCO-8B-Instruct
1,716