Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. (y - x)y' = y - x + 2; y = x + 2√x + 5 When y = x + 2√√x + 5, y' Thus, in terms of x, (y-x)y' = y-x+ 2 = Since the left and right hand sides of the differential equation are equal when x + 2√x + 5 is substituted for y, y = x + 2√x + 5 is a solution. Proceed as in Example 6, by considering simply as a function and give its domain. (Enter your answer using interval notation.) Then by considering as a solution of the differential equation, give at least one interval I of definition. O (-00,-5) O (-10, -5] O (-10, 5) O [-5, 5] O (-5,00)