Question 13

Solving each ODE: - $y_1' = \sin^2 x\, y_1 \Rightarrow y_1 = C_1 e^{\int \sin^2 x\,dx}$ - $y_2' = \cos^2 x\, y_2 \Rightarrow y_2 = C_2 e^{\int \cos^2 x\,dx}$ So $y_1 y_2 = C_1C_2 e^{x + c}$, i.e. $y_1y_2 = K e^{x}$. Using $x=1$: $y_1(1)y_2(1) = 5\cdot\frac13 = \frac53 = Ke$, so $K = \frac{5}{3e}$, giving $y_1y_2 = \frac{5}{3e}e^{x}$. - $y_3' = \left(\frac{2}{x^3} - 1\right)y_3 \Rightarrow \ln y_3 = -\frac{1}{x^2} - x + c$, so $y_3 = C_3 e^{-1/x^2 - x}$. At $x=1$: $C_3 e^{-2} = \frac{3}{5e} \Rightarrow C_3 = \frac{3e}{5}$. Product: $y_1y_2y_3 = \frac{5}{3e}\cdot\frac{3e}{5}\, e^{x}e^{-1/x^2-x} = e^{-1/x^2}$. Limit: $$\lim_{x\to0^+}\frac{e^{-1/x^2} + 2x}{e^{3x}\sin x} = \lim_{x\to 0^+}\left(\frac{e^{-1/x^2}}{x}\cdot\frac{x}{\sin x}\cdot e^{-3x} + \frac{2x}{\sin x}e^{-3x}\right) = 0 + 2 = 2.$$

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