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Take the first and second derivative of the given solution.
Then do a little substitution of y...
@mathlife You should really follow abb0t's advice. Do not expect us to give a solution for you.
If it worked out, close the question. If you are still having trouble, let us know.
there is a neat way to do it with matrices :)
I don't think they should expect ANY answers for any math course above Calculus 1.
@TuringTest HAHAHA i prefer to stay away from that route. Nope. No way.
aw, but it's really pretty cool and fast, plus it works for any set of exponents e^ax, e^bx
@TuringTest No need to make things more complicated... you may have to end up explaining matrix operations to OP.
Can't you use Wronskian too?
Don't worry, I wasn't going to post, you abb0t has described enough no, much easier than the wronskian
no, because you don't get a square matrix
BTW guys I think @mathlife is leaving out some info - I can think of a first-order DE that satisfies this as well as the double-root second-order linear DE we're all thinking of...