The answer is a function AB & BC

$x^{2} - 5x + 6 = 0$ asks which numbers make a statement true, and the answers are $2$ and $3$. $\frac{dy}{dt} = 2y$ asks which functions make a statement true, and the answers include $e^{2t}$, $5e^{2t}$, $-\tfrac{1}{2}e^{2t}$ and infinitely many more.

That difference changes what verification means. There is nothing to solve for and nothing to isolate. You are handed a candidate function, and the job is to test whether substituting it turns the equation into a true statement for every value of the independent variable, not for one value you happen to try.

A useful consequence: you can verify a solution you could never have found. Checking is strictly easier than solving, which is why AP asks for it directly and why it is worth doing on your own work whenever there is time.

A second consequence, which the next section makes into a procedure: the equation has two sides, and both of them depend on the candidate. Substituting into one side and stopping proves nothing at all.

Given a candidate $y$ and an equation such as $\frac{dy}{dt} = 2y$:

1. Compute the left side. Differentiate the candidate. This is the step students skip, and skipping it makes the check vacuous.
2. Compute the right side. Substitute the candidate wherever $y$ appears.
3. Compare the two expressions. If they are identical as functions, the candidate is a solution. If they differ anywhere, it is not.

Test $y = 3e^{2t}$. The left side is $\frac{dy}{dt} = 6e^{2t}$. The right side is $2y = 2 \cdot 3e^{2t} = 6e^{2t}$. Identical, so $y = 3e^{2t}$ is a solution.

Test $y = 2e^{3t}$ in the same equation. The left side is $6e^{3t}$; the right side is $4e^{3t}$. Those are different functions, so it is not a solution, and no value of $t$ except a coincidence will hide that.

Higher-order equations work exactly the same way with more differentiating. For $y'' + 4y = 0$ and $y = \sin(2t)$: $y' = 2\cos(2t)$, $y'' = -4\sin(2t)$, so $y'' + 4y = -4\sin(2t) + 4\sin(2t) = 0$. True for every $t$.

Implicit candidates are no different, they just need implicit differentiation. For $y^{2} = x^{2} + C$, differentiating gives $2y\frac{dy}{dx} = 2x$, so $\frac{dy}{dx} = \frac{x}{y}$. The relation solves that equation without ever being written as a function of $x$.

$\frac{dy}{dt} = 2y$ has a whole family of solutions, $y = Ce^{2t}$, one for each value of $C$. That family is the general solution. Pick a value of $C$ and you have a particular solution, a single function.

An initial value problem supplies a point the solution must pass through, and that point selects one member of the family. Verifying a candidate against an initial value problem therefore has two parts:

1. Does it satisfy the differential equation, checked as above?
2. Does it satisfy the initial condition, checked by substituting the given input?

Both are required. For $\frac{dy}{dt} = 2y$ with $y(0) = 5$: the candidate $y = 5e^{2t}$ passes both, since $y(0) = 5e^{0} = 5$. The candidate $y = 3e^{2t}$ satisfies the equation and gives $y(0) = 3$, so it is a solution of the equation and not of the problem. The candidate $y = 5 + e^{2t}$ passes the initial check at $t = 0$ only by accident of arithmetic and fails the equation, since its derivative $2e^{2t}$ is not $2(5 + e^{2t})$.

Reporting the general solution when a particular one was asked for is the most common way to lose the last mark on these problems. If the prompt hands you a point, the point is there to be used.

Three near-misses, all of which feel like verification and none of which is.

1. Testing one value of $t$. Two different functions can agree at a point, and can even have equal derivatives there. Consider $\frac{dy}{dt} = 2y$ with the candidate $y = 2 + 4t$. At $t = 0$ the left side is $4$ and the right side is $2(2) = 4$: a perfect match. At $t = 0.5$ the left side is still $4$ while the right side is $2(4) = 8$. It satisfies the equation at exactly one instant, which is not what the equation asks.
2. Substituting into the right side only. This produces an expression, and then there is nothing to compare it to. The derivative is half of the statement.
3. Differentiating the wrong thing. In $\frac{dy}{dx} = \frac{x}{y}$ the left side is the derivative of the candidate, not of the right-hand side. Differentiating $\frac{x}{y}$ answers a question nobody asked.

Sign errors deserve their own warning because they survive a sloppy check. $y = e^{-3t}$ has $\frac{dy}{dt} = -3e^{-3t}$, which solves $\frac{dy}{dt} = -3y$ and not $\frac{dy}{dt} = 3y$. Writing the derivative down before comparing is what catches it.

One habit worth building: after producing a solution by any method at all, verify it. The check takes a fraction of the time the solving took, and it is the only self-contained way to know the answer is right.