Answer
$y=\dfrac{2x^{3/2}}{3(1+x)}+\dfrac{c}{1+x}$
Work Step by Step
Here, we have $(1+x) \dfrac{dy}{dx}+y=\sqrt x$ ...(1)
or, $\dfrac{dy}{dx}+\dfrac{y}{1+x}=\dfrac{\sqrt x}{1+x}$
The integrating factor is: $v(x)=e^{\int (1/1+x) dx}=1+x$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$(1+x)\dfrac{dy}{dx}+(1+x)\dfrac{y}{1+x}=(1+x)\dfrac{\sqrt x}{1+x}$
and $\int [(1+x)y)' dx=\int \sqrt x dx$
or, $(1+x)y=\dfrac{2}{3}x^{3/2}+c$
Thus, the general solution is as follows:
$y=\dfrac{2x^{3/2}}{3(1+x)}+\dfrac{c}{1+x}$
