For all $x > 0$, let $y_{1}(x), y_{2}(x), y_{3}(x)$ be the functions satisfying $\frac{dy_{1}}{dx} - (\sin x)^{2}\,y_{1} = 0,\quad y_{1}(1) = 5,$ $\frac{dy_{2}}{dx} - (\cos x)^{2}\,y_{2} = 0,\quad y_{2}(1) = \frac{1}{3},$ $\frac{dy_{3}}{dx} - \left(\frac{2 - x^{3}}{x^{3}}\right) y_{3} = 0,\quad y_{3}(1) = \frac{3}{5e},$ respectively. Then $\lim_{x \to 0^{+}} \frac{y_{1}(x)y_{2}(x)y_{3}(x) + 2x}{e^{3x} \sin x}$ is equal to ________.

For all x > 0, let y₁(x), y₂(x), y₃(x) be the functions satisfying dy₁/dx - (sin x)² y₁ = 0, y₁(1) = 5, dy₂/dx - (cos x)² y₂ = 0, y₂(1) = 1/3, dy₃/dx - ((2 - x³)/x³) y₃ = 0, y₃(1) = 3/5e, respectively. Then lim_(x → 0⁺) (y₁(x)y₂(x)y₃(x) + 2x)/(e³ˣ sin x) is equal to ________.

$\dfrac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1)=5$

$\dfrac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1)=\dfrac13$

$\dfrac{dy_3}{dx} - \left(\dfrac{2-x^3}{x^3}\right)y_3 = 0, \quad y_3(1)=\dfrac{3}{5e}$

$\text{To find: } L = \lim_{x\to 0^+} \dfrac{y_1 y_2 y_3 + 2x}{e^{3x}\sin x}$

$\dfrac{dy_i}{dx} = q_i(x)\,y_i \ \Rightarrow \ \ln y_i = \int q_i(x)\,dx + C_i$

$\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4}, \quad \int \cos^2 x\,dx = \dfrac{x}{2} + \dfrac{\sin 2x}{4}$

$\ln y_1 = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C_1, \quad \ln y_2 = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C_2$

$\ln(y_1 y_2) = x + (C_1 + C_2), \quad y_1(1)y_2(1) = \dfrac53$

$\Rightarrow \ y_1 y_2 = \dfrac{5}{3}\,e^{\,x-1}$

$\ln y_3 = \int \left(\dfrac{2}{x^3} - 1\right)dx = -\dfrac{1}{x^2} - x + C_3$

$y_3 = \dfrac{3}{5}\,e^{\,1 - x - 1/x^2}$

$y_1 y_2 y_3 = \dfrac53 e^{x-1}\cdot\dfrac35 e^{1-x-1/x^2} = e^{-1/x^2}$

$L = \lim_{x\to 0^+}\dfrac{e^{-1/x^2} + 2x}{e^{3x}\sin x} = \lim_{x\to 0^+}\dfrac{e^{-1/x^2}/x + 2}{e^{3x}(\sin x/x)}$

$= \dfrac{0 + 2}{1} = 2$

$\boxed{2}$