EXAMPLE 1 Show that every member of the family of functions y = 1 + cet 1 − cet is a solution of the differential equation y' = 1 2 (y2 − 1). SOLUTION We use the Quotient Rule to differentiate the expression for y: y' = (1 − cet) − (1 + cet)(−cet) (1 − cet)2 = − c2e2t + cet + c2e2t (1 − cet)2 = (1 − cet)2 . The right side of the differential equation becomes 1 2 (y2 − 1) = 1 2 2 − 1 = 1 2 (1 + cet)2 − 2 (1 − cet)2 = 1 2 (1 − cet)2 = (1 − cet)2 . Therefore, for every value of c, the given function is a solution of the differential equation.

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