For all $$x > 0$$, let $$y_1(x)$$, $$y_2(x)$$, and $$y_3(x)$$ 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 ________.

The three linear first-order differential equations have the common form
$$\frac{dy}{dx}-P(x)\,y=0 \; \Longrightarrow \; \frac{dy}{y}=P(x)\,dx \; \Longrightarrow \; y=C\,e^{\int P(x)\,dx}$$

For $$y_1(x):$$
$$P_1(x)=\sin^2 x=\frac{1-\cos 2x}{2}$$
$$\int P_1(x)\,dx=\int\frac{1-\cos 2x}{2}\,dx =\frac{x}{2}-\frac{\sin 2x}{4}$$
Hence $$y_1(x)=C_1\,e^{\,\frac{x}{2}-\frac{\sin 2x}{4}}$$

Using $$y_1(1)=5$$,
$$C_1=5\,e^{-\left(\frac{1}{2}-\frac{\sin 2}{4}\right)}$$
Therefore
$$y_1(x)=5\,\exp\!\Bigl(\!-\!\!\int_{x}^{1}\!\sin^2 t\,dt\Bigr) =5\,e^{\frac{x}{2}-\frac{\sin 2x}{4}}\,e^{-\left(\frac{1}{2}-\frac{\sin 2}{4}\right)}$$

For $$y_2(x):$$
$$P_2(x)=\cos^2 x=\frac{1+\cos 2x}{2}$$
$$\int P_2(x)\,dx=\int\frac{1+\cos 2x}{2}\,dx =\frac{x}{2}+\frac{\sin 2x}{4}$$
So $$y_2(x)=C_2\,e^{\,\frac{x}{2}+\frac{\sin 2x}{4}}$$

Using $$y_2(1)=\dfrac13$$,
$$C_2=\frac13\,e^{-\left(\frac{1}{2}+\frac{\sin 2}{4}\right)}$$
Thus
$$y_2(x)=\frac13\,\exp\!\Bigl(\!-\!\!\int_{x}^{1}\!\cos^2 t\,dt\Bigr) =\frac13\,e^{\frac{x}{2}+\frac{\sin 2x}{4}}\,e^{-\left(\frac{1}{2}+\frac{\sin 2}{4}\right)}$$

For $$y_3(x):$$
$$P_3(x)=\frac{2-x^3}{x^3}=\frac{2}{x^3}-1$$
$$\int P_3(x)\,dx=\int\!\left(\frac{2}{x^3}-1\right)dx =-x^{-2}-x$$
Hence $$y_3(x)=C_3\,e^{-x^{-2}-x}$$

With $$y_3(1)=\dfrac{3}{5e}$$,
$$C_3\,e^{-2}=\frac{3}{5e}\;\Longrightarrow\; C_3=\frac{3e}{5}$$
Therefore
$$y_3(x)=\frac{3e}{5}\,e^{-x^{-2}-x}$$

Behaviour as $$x\to 0^+$$
The exponents in $$y_1(x)$$ and $$y_2(x)$$ approach fixed finite values, so
$$\lim_{x\to 0^+}y_1(x)=A,\qquad \lim_{x\to 0^+}y_2(x)=B$$
where $$A,B$$ are positive constants.

For $$y_3(x)$$ we have the dominant term $$e^{-x^{-2}}$$; thus
$$0\lt y_3(x)\lt K\,e^{-1/x^2}\;\;(\text{for some }K)$$
and consequently
$$y_1(x)\,y_2(x)\,y_3(x)=O\!\bigl(e^{-1/x^2}\bigr)\qquad \bigl(x\to 0^+\bigr)$$
which goes to zero faster than any positive power of $$x$$.

Numerator near $$x=0$$
$$y_1(x)\,y_2(x)\,y_3(x)+2x=2x+o(x)$$

Denominator near $$x=0$$
$$e^{3x}=1+3x+O(x^2),\qquad \sin x=x-\frac{x^{3}}{6}+O(x^{5})$$
Thus
$$e^{3x}\sin x =\bigl(1+3x+O(x^{2})\bigr) \bigl(x-\tfrac{x^{3}}{6}+O(x^{5})\bigr) =x+3x^{2}+O(x^{3})$$

Limit
$$\lim_{x\to 0^+}\frac{y_1(x)\,y_2(x)\,y_3(x)+2x}{e^{3x}\sin x} =\lim_{x\to 0^+}\frac{2x+o(x)}{x+3x^{2}+O(x^{3})} =\lim_{x\to 0^+}\frac{2+o(1)}{1+3x+O(x^{2})}=2$$

Hence the required limit equals 2.