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] I need to solve three differential equations to find ( y_1(x) ), ( y_2(x) ), and ( y_3(x) ), then compute the limit as ( x \to 0^+ ) of the given expression.
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Solve for ( y_1(x) ): The differential equation is ( \frac{dy_1}{dx} - (\sin x)^2 y_1 = 0 ) with ( y_1(1) = 5 ). This is a separable equation. The solution is ( y_1(x) = C_1 e^{\int (\sin x)^2 dx} ). Using the initial condition, we can find ( C_1 ).
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Solve for ( y_2(x) ): The differential equation is ( \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0 ) with ( y_2(1) = 13 ). Similarly, the solution is ( y_2(x) = C_2 e^{\int (\cos x)^2 dx} ). Using the initial condition, we can find ( C_2 ).
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Solve for ( y_3(x) ): The differential equation is ( \frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0 ) with ( y_3(1) = 35e ). The solution is ( y_3(x) = C_3 e^{\int \left(\frac{2 - x^3}{x^3}\right) dx} ). Using the initial condition, we can find ( C_3 ).
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Compute the limit: After finding ( y_1(x) ), ( y_2(x) ), and ( y_3(x) ), substitute them into the expression and compute the limit as ( x \to 0^+ ).
I will use Python to compute the integrals and the limit numerically. [/THOUGHT] [PYTHON] import sympy as sp
x = sp.symbols('x')
Solve for y1(x)
integrand1 = (sp.sin(x))**2 integral1 = sp.integrate(integrand1, x) y1 = sp.exp(integral1) C1 = 5 / y1.subs(x, 1) y1_sol = C1 * sp.exp(integral1)
Solve for y2(x)
integrand2 = (sp.cos(x))**2 integral2 = sp.integrate(integrand2, x) y2 = sp.exp(integral2) C2 = 13 / y2.subs(x, 1) y2_sol = C2 * sp.exp(integral2)
Solve for y3(x)
integrand3 = (2 - x3) / x3 integral3 = sp.integrate(integrand3, x) y3 = sp.exp(integral3) C3 = 35 * sp.exp(1) / y3.subs(x, 1) y3_sol = C3 * sp.exp(integral3)
Compute the limit
numerator = y1_sol * y2_sol * y3_sol + 989 * x denominator = x**3 * sp.exp(1) * sp.sin(x) limit_expr = numerator / denominator limit_val = sp.limit(limit_expr, x, sp.oo, '+')
print(limit_val) [/PYTHON]
[THOUGHT] The Python output shows that the limit evaluates to 0. This is the correct result based on the computation. [/THOUGHT] \boxed{0}