verifying a solution by substitution

Here is a wonderfully reassuring fact about differential equations: even when finding a solution is hard, CHECKING a proposed one is easy and completely mechanical. You do not need to know how the answer was found, or whether the method was legitimate. You just take the candidate function, compute its derivatives, drop them into the equation, and see whether the two sides agree. If they do, it is a solution — full stop.

Precisely, to verify a proposed solution you compute every derivative the equation requires, substitute the function and those derivatives into both sides, simplify, and confirm the equation reduces to a true identity for all x in the interval of interest. Worked example: is y = e^(-x) a solution of y' + y = 0? Compute y' = -e^(-x). Substitute: y' + y = -e^(-x) + e^(-x) = 0. The left side equals the right side (zero) for every x, so yes, it is verified. The check is purely algebra and differentiation — no solving involved.

This skill is more important than it looks. It lets you confirm a textbook answer, catch an arithmetic slip, or test a guessed solution (the method of undetermined coefficients is essentially educated guessing followed by verification). It also clarifies what 'being a solution' really means: not that you derived it correctly, but that it satisfies the equation everywhere. For implicit solutions you do the same thing using implicit differentiation. Whenever you are unsure whether something solves an equation, do not argue — substitute, and let the equation decide.