Subject: Mathematics — Definite Integration
Question: For all $x > 0$, let $y_{1}(x)$, $y_{2}(x)$, and $y_{3}(x)$ be the functions satisfying
$$\frac{dy_{1}}{dx} - \left( \sin x \right)^{2}y_{1} = 0,\quad\quad y_{1}(1) = 5,$$
$$\frac{dy_{2}}{dx} - \left( \cos x \right)^{2}y_{2} = 0,\quad\quad y_{2}(1) = \frac{1}{3},$$
$$\frac{dy_{3}}{dx} - \left( \frac{2 - x^{3}}{x^{3}} \right)y_{3} = 0,\quad\quad y_{3}(1) = \frac{3}{5e},$$
respectively. Then
$\lim_{x \rightarrow 0}\frac{y_{1}(x)y_{2}(x)y_{3}(x) + 2x}{e^{3x}\sin x}$
is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 2
Solution: $\frac{dy_{1}}{dx} = \sin^{2}x\, y_{1} \Rightarrow \frac{dy_{1}}{y_{1}} = \sin^{2}x\, dx\quad\quad\cdots(i)$
$$\frac{dy_{2}}{dx} = \cos^{2}x\, y_{2} \Rightarrow \frac{dy_{2}}{y_{2}} = \cos^{2}x\, dx\quad\quad\cdots(ii)$$
$$\frac{dy_{3}}{dx} = \frac{2 - x^{3}}{x^{3}}\, y_{3} \Rightarrow \frac{dy_{3}}{y_{3}} = \left( \frac{2}{x^{3}} - 1 \right)dx\quad\quad\cdots(iii)$$
From equations $(i) + (ii) + (iii)$,
$$\int\left( \frac{dy_{1}}{y_{1}} + \frac{dy_{2}}{y_{2}} + \frac{dy_{3}}{y_{3}} \right) = \int\left( \sin^{2}x + \cos^{2}x + \frac{2}{x^{3}} - 1 \right)dx + C.$$
$$\log_{e}\left| y_{1}y_{2}y_{3} \right| = 2\int x^{- 3}\, dx + C$$
$$\log_{e}\left| y_{1}y_{2}y_{3} \right| = - \frac{1}{x^{2}} + C.$$
$$\left| y_{1}y_{2}y_{3} \right| = e^{C - \frac{1}{x^{2}}} = c\, e^{- \frac{1}{x^{2}}}.$$
$$\left| y_{1}(1)y_{2}(1)y_{3}(1) \right| = c\, e^{- 1}$$
$$5 \times \frac{1}{3} \times \frac{3}{5e} = \frac{c}{e},\quad\quad c = 1.$$
$$\left| y_{1}(x)y_{2}(x)y_{3}(x) \right| = e^{- \frac{1}{x^{2}}}.$$
$$\Rightarrow y_{1}(x)y_{2}(x)y_{3}(x) = e^{- \frac{1}{x^{2}}}\quad\text{or}\quad - e^{- \frac{1}{x^{2}}}.$$
$$\lim_{x \rightarrow 0}\frac{y_{1}(x)y_{2}(x)y_{3}(x) + 2x}{e^{3x}\sin x} = \frac{2x + e^{- \frac{1}{x^{2}}}}{e^{3x}\sin x} = 2.$$
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