Answer
$y = \sqrt{x}+\frac{C\sqrt{x}}{x}$
Work Step by Step
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation:
$2xy' + y = 2\sqrt{x}$
$y' +(\frac{1}{2x})~y = \frac{\sqrt{x}}{x}$
Let $P(x) = \frac{1}{2x}$
Let $Q(x) = \frac{\sqrt{x}}{x}$
We can find the integrating factor:
$I(x) = e^{\int \frac{1}{2x} ~dx} = e^{0.5~ln(x)} = e^{ln(x^{0.5})} = \sqrt{x} $
We can multiply by the integrating factor:
$\sqrt{x}~y'+\frac{\sqrt{x}}{2x}~y = 1$
$\frac{d}{dx}(\sqrt{x}~y) = 1$
We can integrate both sides of the equation:
$\sqrt{x}~y = \int 1~dx$
$\sqrt{x}~y = x+C$
$y = \sqrt{x}+\frac{C\sqrt{x}}{x}$
