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anonymous
 2 years ago
y''2y'2y=0
anonymous
 2 years ago
y''2y'2y=0

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myininaya
 2 years 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i need detailed answer..tnx
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