证明∑(0->∞)(x^n)/[(n!)^2]满足方程xy''+y'-y=0

Question

证明∑(0->∞)(x^n)/[(n!)^2] 满足方程xy''+y'-y=0

Answer 1

y=∑(0->∞)(x^n)/[(n!)^2]=1+∑(1->∞)(x^n)/[n!*n!]
y'=∑(1->∞)(x^(n-1))*n/[n!*n!]=∑(1->∞)(x^(n-1))/[n!*(n-1)!]
y'=1+∑(2->∞)(x^(n-1))/[n!*(n-1)!]
y''=∑(2->∞)(x^(n-2))*(n-1)/[n!*(n-1)!]
xy''=∑(2->∞)(x^(n-1))/[n!*(n-2)!]=∑(2->∞)(x^(n-1))*(n-1)/[n!*(n-1)!]
xy''+y'=∑(2->∞)(x^(n-1))*(n-1)/[n!*(n-1)!]+1+∑(2->∞)(x^(n-1))/[n!*(n-1)!]
=1+∑(2->∞)(x^(n-1))/[(n-1)!*(n-1)!]
=1+∑(1->∞)(x^(n))/[(n)!*(n)!]
=y
故xy''+y'-y=0