Recent content by jrbigfish

so we end with y=Ce^(1±i√3)t ?? Help me here it has being 12 years after the B.S. is this correct?

so we end with y=Ce^(1±√3)t ?

Ok -1 ± i√3 then what I do?

Ok after I use the quadratic formula (-2+/-((12))^1/2)/2 which solves 0.732i and -2.73i. So the solution is of the form underdamped. So I use the equation y=e^(∝x) (C_1 cosβx+C_2 sinβx). then what?

Ok thank you mag.

Ok the quadratic formula! Yes! But what is this roots of the characteristic equation are r1,2=-a+/-(a^2-b)^1/2 what is this and when I use it?

I hope my posting is not against the rules. The equation is not that difficult but I had being out of school for 20 years. y"+2y'+4y=0 Then I do r^2+2r+4=0 then you can not do (r+2)(r+2)=0 because do not work. can I have some help please?