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anonymous
 5 years ago
Differential equations
anonymous
 5 years ago
Differential equations

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we let \[y= e^{mx}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know that we can get \[(m^2 +3m4)e^{mx}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use characteristic equation: m^2+3m4=0 m=4 and m=1 so y=c1e^(4t)+c2e^t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(d^2+3d4)y=0 d=4,1 y=ae^4x+be^x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and it factors into \[(m1)(m+4)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so since the roots are real and distinct, y = \[ae^{x}+be^{4x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it will be ae^4x+be^x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but what if \[y"10y'+25y = 0 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because your char. equation m^2+3m4=0 has the roots 4,1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you have two same roots m=5 and then y=c1te^5t+c2e^5t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do we proceed the problem? do we let \[y= xe^{mx}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"because your char. equation m^2+3m4=0 has the roots 4,1" dipank this is NOT CORRECT

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i checked........sorry for that........

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0noproblem, i just didn't want to confuse..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the other root I think there was another way to solve it when the roots are equal and real, the formula was different and I think dipank sounds right but I am not sure how to prove it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0long proof..it'd be in your book for sure. circuit theory is the only way i ever wrapped my brain around this concept..a critically damped RLC circuit is modelled by a characteristic equation with two real same roots. Good luck with your proof and further studies though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is a link that might help you understand repeated roots in characteristic equation http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx
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