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find the Differential Equation of the given Family.
Y= Ce^(2x) + e^(2x)
 one year ago
 one year ago
find the Differential Equation of the given Family. Y= Ce^(2x) + e^(2x)
 one year ago
 one year ago

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abb0tBest ResponseYou've already chosen the best response.2
Take the first and second derivative of the given solution.
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Then do a little substitution of y...
 one year ago

vf321Best 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.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
If it worked out, close the question. If you are still having trouble, let us know.
 one year ago

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

abb0tBest ResponseYou've already chosen the best response.2
I don't think they should expect ANY answers for any math course above Calculus 1.
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
@TuringTest HAHAHA i prefer to stay away from that route. Nope. No way.
 one year ago

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

vf321Best 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.
 one year ago

abb0tBest ResponseYou've already chosen the best response.2
Can't you use Wronskian too?
 one year ago

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

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

vf321Best ResponseYou've already chosen the best response.1
BTW 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...
 one year ago
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