Finding Particular Solutions Using Initial Conditions and Separation of Variables AB & BC

An initial condition selects one member of the general family, and the answer is not finished until it is used. For $\frac{dy}{dx} = xy$ with $y(0) = 3$, the family $y = Ce^{x^{2}/2}$ becomes $y = 3e^{x^{2}/2}$. The condition may be applied at the logarithm step, where a positive given value resolves the absolute value and no sign question ever arises, or after solving for $y$; it may not be applied before integrating. When the input is not zero the exponential does not simplify: $\frac{dy}{dx} = 2y$ with $y(1) = 5$ gives $y = 5e^{2x - 2}$, not $5e^{2x}$.

An implicit answer usually carries a sign, and the initial point chooses the branch: $y^{2} = x^{2} + 16$ with $y(0) = -4$ means $y = -\sqrt{x^{2} + 16}$, and a solution never switches branches later. The particular solution also has a DOMAIN, the largest connected interval containing the initial point on which the formula is defined. Solving $\frac{dy}{dx} = y^{2}$ with $y(0) = \frac{1}{2}$ gives $y = \frac{1}{2 - x}$ on $(-\infty, 2)$; the piece beyond the asymptote is a different solution, and denominators, square roots and logarithms are the three things that cut a domain short.