Limit of a Product of Solutions to Differential Equations

Let $Y(x) = y_1(x)y_2(x)y_3(x)$. Summing the logarithmic derivatives of the three given differential equations yields:

$$\frac{Y'(x)}{Y(x)} = \frac{y_1'}{y_1} + \frac{y_2'}{y_2} + \frac{y_3'}{y_3} = \sin^2 x + \cos^2 x + \left(\frac{2-x^3}{x^3}\right) = \frac{2}{x^3}$$

Integrating both sides from $1$ to $x$:

$$\ln\left(\frac{Y(x)}{Y(1)}\right) = \int_1^x \frac{2}{t^3}\,dt = 1 - \frac{1}{x^2}$$

Using the given initial conditions, $Y(1) = y_1(1)y_2(1)y_3(1) = 5 \cdot \frac{1}{3} \cdot \frac{3}{5e} = \frac{1}{e}$, we obtain:

$$Y(x) = e^{-1/x^2}$$

Now, evaluating the limit as $x \to 0^+$:

$$\lim_{x \to 0^+} \frac{y_1(x)y_2(x)y_3(x) + 2x}{e^{3x}\sin x} = \lim_{x \to 0^+} \frac{e^{-1/x^2} + 2x}{e^{3x}\sin x} = \lim_{x \to 0^+} \frac{e^{-1/x^2}}{e^{3x}\sin x} + \lim_{x \to 0^+} \frac{2x}{e^{3x}\sin x} = 0 + 2 = 2$$