各位大佬
n=1,y=[x^(b+1)]/(b+1)+c(1)
n=2,y=[x^(b+2)]/[(b+1)(b+2)]+c(1)x+c(2)
n=3,y=[x^(b+3)]/[(b+1)(b+2)(b+3)+c(1)x²/2+c(2)x+c(3)
……
y=[x^(b+n)/∏(b+m)+∑{c(m)[x^(n-m)]/m!}
m=1~n,c(m)是常数。
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n=1,y=[x^(b+1)]/(b+1)+c(1)
n=2,y=[x^(b+2)]/[(b+1)(b+2)]+c(1)x+c(2)
n=3,y=[x^(b+3)]/[(b+1)(b+2)(b+3)+c(1)x²/2+c(2)x+c(3)
……
y=[x^(b+n)/∏(b+m)+∑{c(m)[x^(n-m)]/m!}
m=1~n,c(m)是常数。