Answer
$$ y= \frac{2}{5}x^{1/2}+\frac{C}{x^2}$$
Work Step by Step
Given $$2 x y^{\prime}+y=2 \sqrt{x}$$
Rewrite the equation
$$ y'+ \frac{1}{2x} y= \frac{1}{\sqrt{x}} $$
Since the equation is linear and $p(x)= \dfrac{2}{x} $ and $q(x)= \dfrac{1}{\sqrt{x}} $, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int (2/x)dx}\\
&=e^{2\ln x}\\
&= x^2
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(x)y&=\int \mu(x)q(x)dx\\
x^2 y&= \int x^2 \dfrac{1}{\sqrt{x}}dx\ \ \ \\
&= \int x^{3/2}dx\\
&= \frac{2}{5}x^{5/2}+C
\end{align*}
Then
$$ y= \frac{2}{5}x^{1/2}+\frac{C}{x^2}$$
