Everything with a $y$ on one side AB & BC

An equation is separable when the right side can be written as a product of something in $x$ alone and something in $y$ alone:

$$\frac{dy}{dx} = g(x) \cdot h(y).$$

Test it by trying to factor. $\frac{dy}{dx} = xy$ is already a product. $\frac{dy}{dx} = y\cos x$ is a product. $\frac{dy}{dx} = \frac{x + 1}{y^{2}}$ is a product, since dividing is multiplying by $\frac{1}{y^{2}}$. And $\frac{dy}{dx} = 3y$ is a product with $g(x) = 3$.

Sums usually do not factor. $\frac{dy}{dx} = x + y$ is not separable, and neither is $\frac{dy}{dx} = x^{2} + y$ or $\frac{dy}{dx} = \sin(x + y)$. Those equations have solutions, and separation is not the way to them. On AP they appear as slope-field or Euler problems for exactly that reason.

A near-miss worth catching: $\frac{dy}{dx} = xy + x$ is separable, because it factors as $x(y + 1)$. Try the factoring before deciding a sum is hopeless.

Take $\frac{dy}{dx} = xy$ all the way through.

1. Separate. $\frac{dy}{y} = x\,dx$. Every $y$ is on the left with $dy$ and every $x$ on the right with $dx$. If a $y$ remains on the right, the separation is not finished and integrating is not yet legal.
2. Integrate both sides. $\int \frac{dy}{y} = \int x\,dx$, giving $\ln|y| = \frac{x^{2}}{2} + C$. One constant, written on the $x$ side. A constant on each side would immediately combine into one, so nothing is lost.
3. Solve for $y$. Exponentiate: $|y| = e^{x^{2}/2 + C} = e^{C} \cdot e^{x^{2}/2}$. The constant has become a factor, because $e^{a + b} = e^{a}e^{b}$.
4. Absorb. $e^{C}$ is just some positive number, and the absolute value contributes a $\pm$, so write $y = Ce^{x^{2}/2}$ with $C$ any nonzero constant. Allowing $C = 0$ recovers the equilibrium solution $y = 0$ that dividing by $y$ removed in step 1.

Check the result by differentiating: $y' = Cx e^{x^{2}/2} = xy$. That check costs one line and catches most errors.

The constant enters at the integration step and nowhere else. Two things follow, and both are examined.

It cannot be added at the end. Writing $y = e^{x^{2}/2} + C$ describes a completely different family: those curves are vertical shifts of one another, while $y = Ce^{x^{2}/2}$ are vertical stretches. Differentiating the wrong one gives $y' = xe^{x^{2}/2}$, which is $x(y - C)$ rather than $xy$.

It cannot be attached before integrating. A $C$ written into the separated equation has nothing to be the constant of, and it will not survive the integration correctly.

Which side to put it on is arbitrary and putting it on both is pointless. $\ln|y| + C_{1} = \frac{x^{2}}{2} + C_{2}$ is the same statement as $\ln|y| = \frac{x^{2}}{2} + C$ with $C = C_{2} - C_{1}$.

One more habit from Unit 6 that pays here: after integrating $\frac{1}{y}$, the answer is $\ln|y|$, bars included. The bars are what let the final constant be negative, which is how solutions below the axis survive.

Not every separable equation ends in an exponential, and the constant behaves differently in each case. Three patterns cover most of what appears.

1. Both sides are powers. $\frac{dy}{dx} = \frac{2x}{y}$ separates to $y\,dy = 2x\,dx$, so $\frac{y^{2}}{2} = x^{2} + C$ and $y^{2} = 2x^{2} + C$. Note that doubling absorbed the old constant into a new one; renaming a constant after multiplying is legitimate and worth saying out loud.
2. A reciprocal on the left. $\frac{dy}{dx} = y^{2}$ gives $-\frac{1}{y} = x + C$, so $y = -\frac{1}{x + C}$. Here the constant sits inside a denominator and cannot be moved out.
3. A quotient of linear factors. $\frac{dy}{dx} = \frac{y}{x}$ gives $\ln|y| = \ln|x| + C$, so $|y| = e^{C}|x|$ and $y = Cx$. The exponential collapses because both sides were logarithms.

Leaving the answer implicit is often acceptable, and sometimes required: $y^{2} = 2x^{2} + C$ is a complete general solution, and solving it for $y$ introduces a sign that nothing yet determines. Topic 7.7 is where an initial condition settles which branch to keep.