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
Very hard differentiation
2009-07-24 11:09 am
回答 (1)
2009-07-24 3:20 pm
✔ 最佳答案
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.
參考: Physics king
收錄日期: 2021-04-19 15:01:01
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