Solve the following initial value problem,
$$\begin{align}
\frac{dy}{dx} &= \exp(2x)(y^2 - 6y + 8)\\
y(-1) &= 3
\end{align}$$This is a separable equation. First we find the general solution following the paradigm.
Separate the variables
$$
\frac{dy}{y^2 - 6y + 8} = \exp(2x) dx
$$
Integrate both sides
$$
(1/2)\log|y - 4| - (1/2)\log|y - 2| = (1/2)\exp(2x) + C
$$
Solve for y (if possible)
We simplify the expression to obtain the general solution
$$
\frac{y - 4}{y - 2} = k \exp(\exp(2x))
$$
Check for singular solutions
We divided by $y^2 - 6y + 8$ which is 0 when $y = 4$ or $y = 2.$ We can check these are both solutions. Now $y = 4$ is already in the general solution (when the constant is 0), but $y = 2$ is a singular solution.