YLL討論網

由 **refa** 於星期六三月 11, 2006 2:30 pm

#ed_op#DIV#ed_cl#                            y'=y  *  (x-1)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/y+3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#之通解為#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                        y + 3ln|y| = 1/3(x-1)#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#+k   ...........(1.3)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#欲滿足y(3)=1，令x= 3 且 y= -1代入方程式(1.3)得#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                       -1 = 1/3(2)#ed_op#SUP#ed_cl#3 #ed_op#/SUP#ed_cl#+ k  ,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故                   k = -11/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此初值問題之解為#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                      y + 3ln|y| = 1/3(x-1)#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl# -11/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而我的問題是為什麼要滿足y(3)=1  ，  令 x=3 且 y= -1呀！   #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                                     #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

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[quote="refa"]#ed_op#DIV#ed_cl#                            y'=y  *  (x-1)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/y+3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#之通解為#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                        y + 3ln\|y\| = 1/3(x-1)#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#+k   ...........(1.3)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#欲滿足y(3)=1，令x= 3 且 y= -1代入方程式(1.3)得#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                       -1 = 1/3(2)#ed_op#SUP#ed_cl#3 #ed_op#/SUP#ed_cl#+ k  ,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故                   k = -11/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此初值問題之解為#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                      y + 3ln\|y\| = 1/3(x-1)#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl# -11/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而我的問題是為什麼要滿足y(3)=1  ，  令 x=3 且 y= -1呀！   #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                                     #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl# [/quote]