Verifying Solutions for Differential Equations AB & BC

A solution of a differential equation is a function, not a number, so verifying one means computing both sides separately and comparing them as expressions for every value of the variable. For $y = 3e^{2t}$ in $\frac{dy}{dt} = 2y$, the left side is $6e^{2t}$ and the right side is $2(3e^{2t}) = 6e^{2t}$, so it is a solution; for $y = 2e^{3t}$ the sides are $6e^{3t}$ and $4e^{3t}$, so it is not. Higher-order equations work the same way with more differentiating, and an implicit relation such as $y^{2} = x^{2} + C$ is checked by implicit differentiation.

The family $y = Ce^{2t}$ is the general solution; one value of $C$ gives a particular solution. An initial value problem needs BOTH checks, the equation and the point: $y = 3e^{2t}$ satisfies $\frac{dy}{dt} = 2y$ but not $y(0) = 5$, so it does not solve that problem. Agreement at one value of $t$ proves nothing, since $y = 2 + 4t$ satisfies $\frac{dy}{dt} = 2y$ at $t = 0$ and nowhere else, and substituting into only the right side leaves nothing to compare against.