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leibnitz theorm
How do i differentiate n times using Leibnitz Theorem:
(1-x^2)y'' - xy' - y = 0
(1-x^2)y'' - xy' - y = 0
Leibnitz's Theorem
(d/dx)^n (uv) = (sum from i = 0 to n) nCi u^(i) v^(n - i)
where u^(i) means the ith derivative of u wrt x, etc.
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(d/dx)^n [(1 - x^2)y'']
= (1 - x^2) y^(n + 2) + n (-2x) y^(n + 1) + nC2 (-2) y^(n)
= (1 - x^2) y^(n + 2) - 2nx y^(n + 1) - n(n - 1) y^(n)
(d/dx)^n (xy') = x y^(n + 1) + n y^(n)
(d/dx)^n y = y^(n)
Combining those three derivatives,
(d/dx)^n [(1 - x^2)y'' - xy' - y]
= (1 - x^2) y^(n + 2) - 2nx y^(n + 1) - n(n - 1) y^(n)
= - x y^(n + 1) - n y^(n)
= - y^(n)
= (1 - x^2) y^(n + 2) - (2n + 1)x y^(n + 1) - (n^2 + 1) y^(n)
(d/dx)^n (uv) = (sum from i = 0 to n) nCi u^(i) v^(n - i)
where u^(i) means the ith derivative of u wrt x, etc.
--
(d/dx)^n [(1 - x^2)y'']
= (1 - x^2) y^(n + 2) + n (-2x) y^(n + 1) + nC2 (-2) y^(n)
= (1 - x^2) y^(n + 2) - 2nx y^(n + 1) - n(n - 1) y^(n)
(d/dx)^n (xy') = x y^(n + 1) + n y^(n)
(d/dx)^n y = y^(n)
Combining those three derivatives,
(d/dx)^n [(1 - x^2)y'' - xy' - y]
= (1 - x^2) y^(n + 2) - 2nx y^(n + 1) - n(n - 1) y^(n)
= - x y^(n + 1) - n y^(n)
= - y^(n)
= (1 - x^2) y^(n + 2) - (2n + 1)x y^(n + 1) - (n^2 + 1) y^(n)
Reply 2
why is it y^(n+2)
Reply 4
thanx 
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