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abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Take the first and second derivative of the given solution.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Then do a little substitution of y...

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1@mathlife You should really follow abb0t's advice. Do not expect us to give a solution for you.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1If it worked out, close the question. If you are still having trouble, let us know.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0there is a neat way to do it with matrices :)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2I don't think they should expect ANY answers for any math course above Calculus 1.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2@TuringTest HAHAHA i prefer to stay away from that route. Nope. No way.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0aw, but it's really pretty cool and fast, plus it works for any set of exponents e^ax, e^bx

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1@TuringTest No need to make things more complicated... you may have to end up explaining matrix operations to OP.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2Can't you use Wronskian too?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0Don't worry, I wasn't going to post, you abb0t has described enough no, much easier than the wronskian

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0no, because you don't get a square matrix

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1BTW guys I think @mathlife is leaving out some info  I can think of a firstorder DE that satisfies this as well as the doubleroot secondorder linear DE we're all thinking of...
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