One curve out of infinitely many AB & BC

Solve $\frac{dy}{dx} = xy$ with $y(0) = 3$. The general solution from Topic 7.6 is $y = Ce^{x^{2}/2}$. Substitute the point: $3 = Ce^{0} = C$, so the particular solution is

$$y = 3e^{x^{2}/2}.$$

That is the whole procedure, and the mark is lost by stopping at the line above it. A response that reports $y = Ce^{x^{2}/2}$ has answered a question about the equation and not the question that was asked, which supplied a point precisely so that one curve could be named.

Two checks worth running on the finished answer. Differentiate it and confirm the equation holds; substitute the initial input and confirm the required output comes back. Both together take under a minute and they catch the constant landing in the wrong place.

The condition can be used at either of two moments, and one of them is easier.

1. At the logarithm step. From $\ln|y| = \frac{x^{2}}{2} + C$, substitute $x = 0$ and $y = 3$ to get $\ln 3 = C$. Then $y = e^{x^{2}/2 + \ln 3} = 3e^{x^{2}/2}$. Since the given value is positive, the absolute value resolves itself and no $\pm$ ever appears.
2. After solving for $y$. From $y = Ce^{x^{2}/2}$, substitute to get $C = 3$. Same answer, one step later.

What is not allowed is applying the condition before integrating. A point is a fact about the solution, not about the differential equation, and substituting it into the separated form produces numbers where variables need to be.

When the initial input is not zero, the exponential does not simplify away. For $\frac{dy}{dx} = 2y$ with $y(1) = 5$: $y = Ce^{2x}$ gives $5 = Ce^{2}$, so $C = 5e^{-2}$ and $y = 5e^{2x - 2}$. Reporting $y = 5e^{2x}$ treats the condition as though it had been given at $x = 0$, and that answer passes through $(0, 5)$ instead of $(1, 5)$.

Separation often produces an implicit answer whose explicit form carries a sign. Solving $\frac{dy}{dx} = \frac{x}{y}$ gives $y^{2} = x^{2} + C$, and taking square roots gives $y = \pm\sqrt{x^{2} + C}$: two curves, one above the axis and one below.

The initial condition chooses. With $y(0) = -4$: substituting into the implicit form gives $16 = 0 + C$, so $y^{2} = x^{2} + 16$. The point has a negative $y$, so the solution is $y = -\sqrt{x^{2} + 16}$, the lower branch, and the upper one is a different solution of the same equation.

The same question arrives disguised as an absolute value. From $\ln|y| = kt + C$ with $y(0) = -8$, the solution stays negative for all $t$, since a solution cannot cross the equilibrium $y = 0$. So $|y| = -y$ throughout and the answer is $y = -8e^{kt}$.

A solution never switches branches partway along, because switching would require passing through the place where the branches meet, and that is either an equilibrium or a point where the derivative fails to exist. Decide the branch once, at the initial point, and it holds.

A particular solution is not automatically defined for every real number. Its domain is the largest connected interval containing the initial point on which the formula makes sense. This is quietly examined and rarely taught.

Solve $\frac{dy}{dx} = y^{2}$ with $y(0) = \frac{1}{2}$. Separating gives $-\frac{1}{y} = x + C$, and the condition gives $C = -2$, so

$$y = \frac{1}{2 - x}.$$

This formula blows up at $x = 2$. The initial point sits at $x = 0$, which is to the left of that, so the particular solution is $y = \frac{1}{2 - x}$ on $(-\infty, 2)$. The piece of the same formula with $x > 2$ is a solution of the differential equation as well, and it is a different solution: no curve joins them, since nothing passes through the asymptote.

Three things that cut a domain short, all worth checking:

1. A denominator reaching zero, as above.
2. A square root turning negative, as in $y = \sqrt{x^{2} - 9}$, which is undefined between $-3$ and $3$.
3. A logarithm's argument reaching zero, which happens where a solution meets an equilibrium.

Report the interval, not the whole real line, whenever one of these is present.