Question 13

The differential equations are of the form $\frac{dy}{dx} = P(x)y \implies y = C e^{\int P(x) dx}$. Let $Y(x) = y_1(x)y_2(x)y_3(x)$. Then $\frac{dY}{dx} = y_1'y_2y_3 + y_1y_2'y_3 + y_1y_2y_3' = (\sin^2 x)y_1y_2y_3 + (\cos^2 x)y_1y_2y_3 + (\frac{2-x^3}{x^3})y_1y_2y_3$ $\frac{dY}{dx} = Y(\sin^2 x + \cos^2 x + \frac{2}{x^3} - 1) = Y(1 + \frac{2}{x^3} - 1) = \frac{2}{x^3} Y$ Integrating: $\int \frac{dY}{Y} = \int \frac{2}{x^3} dx \implies \ln Y = -\frac{1}{x^2} + C \implies Y(x) = k e^{-1/x^2}$. Given $y_1(1)y_2(1)y_3(1) = 5 \cdot \frac{1}{3} \cdot \frac{3}{5e} = \frac{1}{e}$. At $x=1$, $Y(1) = k e^{-1} = \frac{1}{e} \implies k = 1$. So $y_1(x)y_2(x)y_3(x) = e^{-1/x^2}$. Now evaluate the limit: $\lim_{x \to 0^+} \frac{e^{-1/x^2} + 2x}{e^{3x} \sin x}$ As $x \to 0^+$, $e^{-1/x^2} \to 0$ extremely fast (faster than any power of $x$). Limit = $\lim_{x \to 0^+} \left( \frac{e^{-1/x^2}}{e^{3x} \sin x} + \frac{2x}{e^{3x} \sin x} \right)$ The first term is $0$. The second term is $\lim_{x \to 0^+} \frac{2x}{(1)(x)} = 2$. Thus, the limit is $0 + 2 = 2$.