(E) (A), (B), (C), and (D) are all true. (E) is false: see Answer 9.
(A) Integrating yields , where we have
replaced the arbitrary constant 2C by C′.
(C) For initial point (−2,1), x^2 − y^2 = 3. Rewriting the d.e. y dy = x dx as
reveals that the derivative does not exist when y = 0, which occurs
at . Since the particular solution must be differentiable in an
interval containing x = −2, the domain is .
(E) We separate variables. , so ln . The initial point
yields ln ; hence c = −2. With y > 0, the particular solution is ln
, or .
(C) We separate variables. , so −e−y = x + c. The particular
solution is −e−y = x − 2.
(B) The general solution is ; when x = 4
yields C = 0.
(E) Since , it follows that
ln y = ln x + C or ln y = ln x + ln k;
so y = kx.
