- #1
doey
- 27
- 0
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.
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.
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