Answer
$y = \frac{2}{3}~\sqrt{x}+\frac{C}{x}$
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:
$xy' + y = \sqrt{x}$
$y' +(\frac{1}{x})~y = \frac{\sqrt{x}}{x}$
Let $P(x) = \frac{1}{x}$
Let $Q(x) = \frac{\sqrt{x}}{x}$
We can find the integrating factor:
$I(x) = e^{\int \frac{1}{x} ~dx} = e^{ln(x)} = x$
We can multiply by the integrating factor:
$x~y'+y = \sqrt{x}$
$\frac{d}{dx}(x~y) = \sqrt{x}$
We can integrate both sides of the equation:
$x~y = \int \sqrt{x}~dx$
$x~y = \frac{2}{3}~x^{3/2}+C$
$y = \frac{2}{3}~\sqrt{x}+\frac{C}{x}$
