For all , let , , and be the functions satisfying\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1) = 5,$$$$\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1) = 13,$$$$\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0, \quad y_3(1) = 35e,respectively. Find the value of\
[THOUGHT] The DEs are separable. The product y1 y2 y3 = 51335*e * exp(∫_1^x [sin²t + cos²t + (2-t³)/t³] dt) = 2275 e * exp(∫_1^x 2 t^{-3} dt) = 2275 e * exp([-1/t²]_1^x) = 2275 e * exp(-1/x² + 1) = 2275 e² exp(-1/x²).
The expression is [2275 e² exp(-1/x²) + 989 x] / (x³ e sin x).
As x→0+, exp(-1/x²) → 0 faster than any power of x grows, so numerator ∼ 989 x, denominator ∼ e x⁴, ratio ∼ 989/(e x³) → +∞.
To confirm with sympy, compute the limit numerically or symbolically. [/THOUGHT] [PYTHON] import sympy as sp
x = sp.symbols('x', positive=True) e = sp.E product = 2275 * e2 * sp.exp(-1/x2) num = product + 989 * x den = x**3 * e * sp.sin(x) expr = num / den lim = sp.limit(expr, x, 0, dir='+') print(lim)
Also check numerical at small x
print(float(expr.subs(x, 0.01))) print(float(expr.subs(x, 0.001))) [/PYTHON]
[THOUGHT] The Python output confirms the limit is oo (positive infinity). This is the ground truth value of the limit. [/THOUGHT] \boxed{\infty}
Explanation
The provided AI-generated answer concludes that the limit is (\infty), based on the Python output showing "oo" from SymPy's limit computation. The reasoning follows these steps:
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Solve the differential equations: Each equation is of the form ( \frac{dy}{dx} = p(x) y ), separable as ( \frac{dy}{y} = p(x) , dx ). The solutions are ( y_1(x) = 5 \exp\left( \int_1^x \sin^2 t , dt \right) ), ( y_2(x) = 13 \exp\left( \int_1^x \cos^2 t , dt \right) ), and ( y_3(x) = 35e \exp\left( \int_1^x \frac{2 - t^3}{t^3} , dt \right) ).
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Compute the product ( y_1(x) y_2(x) y_3(x) ): Note that ( \sin^2 t + \cos^2 t = 1 ), so ( \int_1^x (\sin^2 t + \cos^2 t) , dt = x - 1 ). For ( y_3 ), ( p_3(t) = \frac{2}{t^3} - 1 ), so ( \int_1^x p_3(t) , dt = \left[ -\frac{1}{t^2} - t \right]_1^x = -\frac{1}{x^2} - x + 2 ). Thus, ( y_1 y_2 y_3 = 5 \cdot 13 \cdot 35e \cdot \exp(x - 1) \cdot \exp\left( -\frac{1}{x^2} - x + 2 \right) = 2275 e^2 \exp\left( -\frac{1}{x^2} \right) ).
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Analyze the numerator: As ( x \to 0^+ ), ( \exp\left( -\frac{1}{x^2} \right) \to 0 ) faster than any polynomial, making ( y_1 y_2 y_3 \approx 0 ). The numerator simplifies to ( 989x + o(x) \sim 989x ).
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Analyze the denominator: ( x^3 e \sin x \sim x^3 e x = e x^4 ) as ( x \to 0^+ ), since ( \sin x \sim x ).
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Evaluate the limit: The expression becomes ( \frac{989x}{e x^4} = \frac{989}{e x^3} \to +\infty ) as ( x \to 0^+ ).
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Verify with Python: The code computes ( \lim_{x \to 0^+} \frac{2275 e^2 \exp(-1/x^2) + 989x}{x^3 e \sin x} ), yielding "oo". Numerical checks at ( x = 0.01 ) and ( x = 0.001 ) show values increasing to approximately ( 3.64 \times 10^8 ) and ( 3.64 \times 10^{11} ), confirming divergence.
This aligns with the Python output as ground truth, establishing the limit as ( +\infty ).