Verify that the following functions are solutions to the given differential equation. y=2ex−x−1 solves y′=y+x

${\displaystyle y=2e^x-x-1\text{ solves }y\prime =y+x}$
${\displaystyle y=2e^x-x-1}$ (1)
${\displaystyle \Rightarrow y^{\prime }=\frac{d}{dx}(2e^x-x-1)}$
${\displaystyle =2e^x-1+0}$ (exponential diff. and general power rule)
${\displaystyle =\overbrace{(2e^x-x-1)}+x(\because y\text{ from}\left(1\right))}$ (reforming)
$=y+x$ (satisfying the differental equation)
Result
${\displaystyle y^{\prime }=\frac{d}{dx}(2e^x-x-1)=2e^x-1+0}$ (exponential diff. and general power rule)
${\displaystyle =(2e^x-x-1)+x=y+x}$ (satisfying the differental equation)