ay'' + by' + cy = 0 1) for 2 distinct roots,the solution should be y = Ae^m1x + Be^m2x 2) for Real and equal roots should be y = e^mx(A + Bx) that is it. So if same roots as the 2nd one,one solution is y1(x)=Ae^mx, what about the second solution? there is a second solution given by y2(x)=Bxe^mx HERE IS THE PROBLEM!!! WHY THERE EXISTED AN xTHERE????!!! is this should be y2(x)=Be^mx also? because looks same as distinct root solution, and we can just put the 2 roots in the same just act like there have 2 distinct roots but are the same.