Question:

Solve each ODE (linear, separable): $y_1$: $\dfrac{dy_1}{y_1} = (\sin x)^2\,dx$, so $\ln y_1 = \int_1^x \sin^2 t\,dt + \ln 5$. $y_1(x) = 5\exp\!\left(\int_1^x \sin^2 t\,dt\right)$. As $x \to 0^+$: $\int_1^x \sin^2 t\,dt = -\int_x^1 \sin^2 t\,dt \to -\int_0^1 \sin^2 t\,dt$. Let $C_1 = \int_0^1 \sin^2 t\,dt$. Then $y_1(0^+) = 5e^{-C_1}$. $y_2(x) = \dfrac{1}{3}\exp\!\left(\int_1^x \cos^2 t\,dt\right)$. Similarly $y_2(0^+) = \dfrac{1}{3}e^{-C_2}$ where $C_2=\int_0^1\cos^2 t\,dt$. $C_1+C_2 = \int_0^1(\sin^2 t+\cos^2 t)\,dt = 1$, so $e^{-(C_1+C_2)}=e^{-1}$. $y_3$: $\dfrac{dy_3}{y_3} = \dfrac{2-x^3}{x^3}dx = \left(\dfrac{2}{x^3}-1\right)dx$. $\ln y_3 = -\dfrac{1}{x^2} - x + C$. At $x=1$: $\ln(3/5e) = -1-1+C$, so $C = \ln(3/5e)+2 = \ln(3/5)+1$. $\ln y_3(x) = -\dfrac{1}{x^2}-x+\ln(3/5)+1$. $y_3(x) = \dfrac{3}{5}\cdot e^{1-x-1/x^2}$. As $x\to0^+$: $e^{-1/x^2}\to 0$, so $y_3(x)\to 0$ but we need the product behaviour. $y_1(x)y_2(x) = \dfrac{5}{3}\exp\!\left(\int_1^x(\sin^2t+\cos^2t)dt\right) = \dfrac{5}{3}\exp(x-1) = \dfrac{5}{3}e^{x-1}$. So $y_1 y_2 y_3 = \dfrac{5}{3}e^{x-1}\cdot\dfrac{3}{5}e^{1-x-1/x^2} = e^{-1/x^2}$. As $x\to0^+$: $e^{-1/x^2}\to0$ faster than any power of $x$, so the numerator $\to 0+0=0$ and denominator $\to 0$. $\lim_{x\to0^+}\dfrac{e^{-1/x^2}+2x}{e^{3x}\sin x}$. Since $e^{-1/x^2}=o(x^n)$ for all $n$: $\approx \dfrac{2x}{x} = 2$ (since $e^{3x}\sin x \approx x$ as $x\to0$). More precisely: $\lim = \lim_{x\to0^+}\dfrac{2x}{e^{3x}\sin x} = \lim_{x\to0^+}\dfrac{2x}{x} = 2$.