Use

$F^{\text{'}}(x)=\frac{d}{dx}{\displaystyle \underset{a}{\overset{x}{}}}g(t)dt=g(x)$

$to$ verify that the indicated function $is$ a solution $of$ the given differential equation. Assume $an$ appropriate interval I $of$ definition $of$ the solution.

$2x\frac{dy}{dx}-y=2xcos(x)$;$,y=\sqrt[\phantom{2}]{}n{t}_4^x\frac{cos(t)}{\sqrt[\phantom{2}]{t}}dt$

When $y=\sqrt[\phantom{2}]{}n{t}_4^x\frac{cos(t)}{\sqrt[\phantom{2}]{t}}dt,$

$\frac{dy}{dx}=(,){\displaystyle \underset{4}{\overset{xcos(t)}{}}}\frac{?}{\sqrt[\phantom{2}]{t}}dt+$

Thus, $in$ terms $ofx,$

$2x\frac{dy}{dx}-y=(1){\displaystyle \underset{4}{\overset{xcos(t)}{}}}\frac{1}{\sqrt[\phantom{2}]{t}}dt+-\sqrt[\phantom{2}]{}n{t}_4^{xcos(t)}\frac{1}{\sqrt[\phantom{2}]{t}}dt$

$=$