$ \begin{aligned} & \frac{d y_1}{d x}-\left(\sin ^2 x\right) y_1=0 \\ & \Rightarrow \int \frac{d y_1}{y_1}=\int \sin ^2 x d x \\ & \Rightarrow \int \frac{d y_1}{y_1}=\int \frac{1-\cos 2 x}{2} d x \\ & \Rightarrow \ln \left|y_1\right|=\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)+C_1 \\ & \Rightarrow y_1=e^{\frac{1}{2}}\left(x-\frac{\sin 2 x}{2}\right)+C_1 \\ & \because y_1(1)=5 \\ & \Rightarrow C_1=\ln 5-\frac{1}{2}+\frac{\sin 2}{4} \\ & \Rightarrow y_1=e^{\frac{1}{2}}\left(x-\frac{\sin 2 x}{2}\right)+\ln 5-\frac{1}{2}+\frac{\sin 2}{4} \ldots \text { (i) } \\ & \text { and } \frac{d y_2}{d x}=\left(\cos ^2 x\right) y_2 \\ & \Rightarrow \int \frac{d y_2}{y_2}=\int \cos ^2 x d x=\int \frac{1+\cos 2 x}{2} d x \\ & \Rightarrow \ln \left|y_2\right|=\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)+C_2 \end{aligned} $
$ \begin{aligned} & \because y_2(1)=\frac{1}{3} \\ & \Rightarrow C_2=-\ln 3-\frac{1}{2}-\frac{\sin 2}{4} \\ & \Rightarrow y_2=e^{\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)-\ln 3-\frac{1}{2}-\frac{\sin 2}{4}} \text { …(2) } \\ & \text { and } \frac{d y_3}{d x}=\left(\frac{2-x^3}{x^3}\right) y_3 \\ & \Rightarrow \int \frac{d y_3}{y_3}=\int\left(\frac{2}{x^3}-1\right) d x \\ & \Rightarrow \ln \left|y_3\right|=-\frac{1}{x^2}-x+C_3 \\ & \because y_3(1)=\frac{3}{5 e} \end{aligned} $
$ \begin{aligned} & \Rightarrow C_3=1+\ln 3-\ln 5 \\ & \Rightarrow y_3=e^{\frac{1}{x^2}-x+1+\ln 3-\ln 5} \text { ...(3) } \end{aligned} $
From eq ${ }^n(1)$, (2) and (3)
$ \begin{aligned} & y_1(x) y_2(x) y_3(x)=e^{-\frac{1}{x^2}} \\ & \therefore \lim _{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x)+2 x}{e^{3 x} \sin x} \\ & =\lim _{x \rightarrow 0^{+}} \frac{e^{\frac{-1}{x^2}}+2 x}{e^{3 x} \sin x} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x} e^{\frac{-1}{x^2}}+2}{e^{3 x}\left(\frac{\sin x}{x}\right)} \\ & =\frac{0+2}{1 \times 1}=2 \end{aligned} $