1. Given the function $y=y_{1}+y_{2}$, where $y_{2}$ is a direct proportion function of $x$, and $y_{2}$ is an inverse proportion function of $x$, and when $x=2$, $y=8$, and when $x=4$, $y=13$. Then the function expression of $y$ with respect to $x$ is $\qquad$ .
=. 1. $y=3 x+\frac{4}{x}$.
Let $y_{1}=k_{1} x, y_{2}=\frac{k_{2}}{x}$. From the given $8=2 k_{1}+\frac{k_{2}}{2}$ and $13=4 k_{1}+\frac{k_{2}}{4}$, we solve to get $k_{1}=3, k_{2}=4$.