Verify that the indicated function $y=(x)$ is an explicit solution of the given first$-$order differential equation.

$(y-x)y^{\text{'}}=y-x+2$;$,y=x+2\sqrt[\phantom{2}]{x+6}$

When $y=x+2\sqrt[\phantom{2}]{x+6}$

$y^{\text{'}}=$

Thus, in terms of $x,$

$(y-x)y^{\text{'}}=$

$y-x+2=$

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Since the left and right hand sides of the differential equation are equal when $x+2\sqrt[\phantom{2}]{x+6}$ is substituted for $y,y=x+2\sqrt[\phantom{2}]{x+6}$ is a solution. Proceed as in Example $6,$ by considering $$ simply as a function and give its domain. $($Enter your answer using interval notation.$)$

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