
#1
Aug411, 03:54 AM

P: 168

Hello experts!
As we know that there are 3 different general solutions of an ordinary differential equation depends on that what type of roots we've gotten in the solution, listed below. 1) y(t)=c_{1}e^{m1t}+c_{2}e^{m2t} 2) y(t)=c_{1}e^{mt}+c_{2}te^{mt} 3) y(t)=c_{1}e^{u}cos(v)+c_{2}sin(v) Now 1st solution is used when discriminant i.e. D>0, 2nd is used when D=0 and 3rd is used when D<0 But here is a question, y^{''}+6y^{'}+9y=0 By solving we get roots, m_{1}=3 and m_{2}=3 While discriminant is 30 & book has used 2nd solution to solve it, if D<0 it must use 3rd solution. How do we choose correct solution? Does solution is according to nature of roots(discriminant) or m_{1} & m_{2}? Thanks for your contribution. 



#2
Aug411, 05:22 AM

P: 25

Discriminant is not 30, it is 6^24.1.9=0




#3
Aug411, 06:14 AM

P: 168

Ok.
Thanks to light me the right way. 


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