Find the singular points of this differential equation and classify them as

Z

zan81289

New member
regular or irregular? x^2(x-2)y"+(5x-7)y'+2(3+5x^2)y=0

I'm getting it all right except for the classification of one of the singular points. Can anyone help me with this. (Answer from book: 0,regular 2,regular)
Thanks Ted, that's what I got as well, I guess its just a mistake in the answer key.
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
as typed 2 is regular but 0 is irregular as lim { x---> 0 } [ x(5x-7) / x²] does not exist...that is 0 is a pole of order 2 for p_1 (x) =(5x-7) / (x² [x-2]) and 2 is a pole of order 1
 
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