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ton
y''-2y'-2y=0
If you have a_[n]*y^[n]+a_[n-1]*y^[n-1]+a_[n-2]*y^[n-2]+....+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_[n-1]*r^(n-1)+....+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.
i need detailed answer..tnx