Problem Analysis #19

The differential equations and initial conditions are given as follows:

• For (y_1): (\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0), with (y_1(1) = 5).
• For (y_2): (\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0), with (y_2(1) = 13).
• For (y_3): (\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0), with (y_3(1) = 35e).

Solving each equation:

• (y_1(x) = 5 \exp\left( \int_1^x (\sin t)^2 dt \right) = 5 \exp\left( \frac{x}{2} - \frac{\sin 2x}{4} - \frac{1}{2} + \frac{\sin 2}{4} \right)).
• (y_2(x) = 13 \exp\left( \int_1^x (\cos t)^2 dt \right) = 13 \exp\left( \frac{x}{2} + \frac{\sin 2x}{4} - \frac{1}{2} - \frac{\sin 2}{4} \right)).
• (y_3(x) = 35e \exp\left( \int_1^x \left( \frac{2}{t^3} - 1 \right) dt \right) = 35e \exp\left( -\frac{1}{x^2} - x + 2 \right)).

The product (y_1(x) y_2(x) y_3(x)) is:

[ y_1 y_2 y_3 = 5 \cdot 13 \cdot 35e \cdot \exp\left( \int_1^x (\sin t)^2 dt + \int_1^x (\cos t)^2 dt + \int_1^x \left( \frac{2}{t^3} - 1 \right) dt \right). ]

The sum of the integrals is:

[ \int_1^x \left( (\sin t)^2 + (\cos t)^2 \right) dt + \int_1^x \left( \frac{2}{t^3} - 1 \right) dt = \int_1^x 1 dt + \int_1^x \left( \frac{2}{t^3} - 1 \right) dt = (x - 1) + \left[ -\frac{1}{t^2} - t \right]_1^x = (x - 1) + \left( -\frac{1}{x^2} - x + 2 \right) = 1 - \frac{1}{x^2}. ]

Thus,

[ y_1 y_2 y_3 = 2275e \cdot \exp\left(1 - \frac{1}{x^2}\right) = 2275e^2 \exp\left(-\frac{1}{x^2}\right). ]

The numerator is:

[ y_1 y_2 y_3 + 989x = 2275e^2 \exp\left(-\frac{1}{x^2}\right) + 989x. ]

The denominator is (x^3 e \sin x). As (x \to 0^+):

• (\exp\left(-\frac{1}{x^2}\right) \to 0) faster than any polynomial decay.
• (\sin x \sim x), so (x^3 e \sin x \sim x^4 e).

The dominant term in the numerator as (x \to 0^+) is (989x), since (2275e^2 \exp\left(-\frac{1}{x^2}\right)) decays faster than any power of (x). Thus, the fraction behaves as:

[ \frac{989x}{x^4 e} = \frac{989}{e x^3} \to +\infty. ]

Therefore, the limit is (+\infty).

\boxed{\infty}

[THOUGHT] We are required to use the Python tool at least once. We will compute the expression for a small positive x (0.001) to observe the behavior. [/THOUGHT] [PYTHON] import math

x = 0.001 integral_sin2 = (x/2 - math.sin(2*x)/4) - (1/2 - math.sin(2)/4) y1 = 5 * math.exp(integral_sin2)

integral_cos2 = (x/2 + math.sin(2*x)/4) - (1/2 + math.sin(2)/4) y2 = 13 * math.exp(integral_cos2)

y3 = 35 * math.e * math.exp(-1/(x**2) - x + 2)

product = y1 * y2 * y3 numerator = product + 989 * x denominator = x**3 * math.e * math.sin(x) expression = numerator / denominator

print(expression) [/PYTHON]

\boxed{\infty}