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, 1 y' = 1 + √x + 5 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 o 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.(-10, 5) O (-∞, -5) Ⓒ (-5,00) O (-10, -5] O [-5, 5]

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