Answer
$y= \frac{2}{3}\sqrt x+\frac{C}{x}$
Work Step by Step
Given $$x y^{\prime}+y=\sqrt{x}$$
Rewrite the equation
$$ y'+ \frac{1}{x} y= \frac{1}{\sqrt{x}} $$
Since the equation is linear and $p(x)= \dfrac{1}{x} $ and $q(x)= \dfrac{1}{\sqrt{x}} $, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int (1/x)dx}\\
&=e^{\ln x}\\
&= x
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
x y&= \int x \dfrac{1}{\sqrt{x}}dx\ \ \ \\
&= \int \sqrt{x} dx\\
&= \frac{2}{3}x^{3/2}+C
\end{align*}
Then
$$ y= \frac{2}{3}\sqrt x+\frac{C}{x}$$
