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ton
 one year ago
y''2y'2y=0
ton
 one year ago
y''2y'2y=0

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.1If you have a_[n]*y^[n]+a_[n1]*y^[n1]+a_[n2]*y^[n2]+....+a_[0]*y=0 where the a's are constants and the number in the ^[ ] is which derivative we are referring to, then the first step is to solve a_[n]*r^n+a_[n1]*r^(n1)+....+a_[0]=0. If r is real and say r=pm a, then your solution will be y=c1*e^(at)+c2*e^(bt). If r is complex (or just includes the imaginary part), then your answer will involve cos(x) and sin(x) instead of exponentials.

ton
 one year ago
Best ResponseYou've already chosen the best response.0i need detailed answer..tnx
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