(C) The general solution is y = 3 ln|x^2 − 4| + C. The differential equation
reveals that the derivative does not exist for x = ±2. The
particular solution must be differentiable in an interval containing the
initial value x = −1, so the domain is −2 < x < 2.
(A) The solution curve shown is y = ln x, so the differential equation is
.
(D).
(C) The equations may be rewritten as = sin u and y = 1 − 2 sin^2 u,