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

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

vf321
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0there is a neat way to do it with matrices :)

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

TuringTest
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.2Can't you use Wronskian too?

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

vf321
 one year 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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