Answer
y is a solution
Work Step by Step
Find $y^{\prime} $ and $ y^{\prime\prime}$
$y=2(\ln x)+3$
$y^{\prime}=2\displaystyle \cdot\frac{1}{x}+0= \frac{2}{x}=(2x^{-1})$
$y^{\prime\prime}=(2x^{-1})^{\prime}=2(-x^{-2})=-\displaystyle \frac{2}{x^{2}}$
Substitute and see if the differential equation is satisfied
$xy^{\prime\prime}+y^{\prime}=x(-\displaystyle \frac{2}{x^{2}})+\frac{2}{x}=-\frac{2}{x}+\frac{2}{x}=0$
It is, so,
y is a solution.
