Verify that the indicated function y = ϕ(x) is an explicit solution of the given first-order differential equation. (y − x)y' = y − x + 18; y = x + 6 √(x + 4) When y = x + 6√(x + 4) y'=
Thus in terms of x, (y-x)y'=
y-x+18=

If y = x + 6√(x + 4), then

y' = 1 + 3/√(x + 4)

Substituting y and y' into the DE gives

(y - x) y' = (x + 6√(x + 4) - x) (1 + 3/√(x + 4))

… = 6√(x + 4) (1 + 3/√(x + 4))

… = 6√(x + 4) + 18

on the left side, while on the right you get

y - x + 18 = x + 6√(x + 4) - x + 18

… = 6√(x + 4) + 18

so both sides match and the given function is indeed a solution to the DE.

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