close
標題:
Very hard differentiation
發問:
If y^(0) = e^(x^2) , y^(n) = differentiates y^(0) n times for all positive integers n , prove that y^(n+1) - 2xy^(n) - 2ny^(n-1) = 0
此文章來自奇摩知識+如有不便請留言告知
最佳解答:
y(0) = ex^2 Differentiate both sides with respect to x. y(1) = 2x ex^2 = 2xy(0) Differentiate both sides with respect to x by n times, by Leibniz's Theorem, y(n+1) = 2xy(n) + 2nC1y(n-1) Therefore, y(n+1) - 2xy(n) - 2ny(n-1) = 0 for all positive integers n.
其他解答:
文章標籤
全站熱搜
留言列表
