Ernstfalld

2021-10-24

Verify that the following functions are solutions to the given differential equation. $y=2{e}^{x}-x-1$ solves $y\prime =y+x$

### Answer & Explanation

cheekabooy

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

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