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प्रश्न
If y=sin[log(x2+2x+1)] then prove that (x+1)2yn+2 +(2n +1)(x+ 1)yn+1 + (n2+4)yn=0.
बेरीज
उत्तर
We have,
y=sin [log(x2+2x+1)]
Diff with x
y1 = cos [log(x2+2x+1)] × `1/(x^2+2x+1)xx(2x+2)`
y1= cos [log(x2+2x+1)] × `(2(x+1))/(x+1)^2`
y1= cos [log(x2+2x+1)] × `2/(x+1)`
(x + 1) y1=2 cos [log(x2+2x+1)]
Diff with x again,
(x + 1) y2 + y1 = - 2 sin [log(x2+2x+1)] × `x/(x^2+2x+1)`× (2x + 2)
(x + 1) y2 + y1 = - 2 sin [log(x2+2x+1)] × `(2(x+1))/(x+1)^2`
(𝑥 + 1)2 y2 + (x + 1)y1 = - 4 sin [log(x2+2x+1)]
(𝑥 + 1)2 y2 + (x + 1)y1 = - 4y
By Leibnitz Theorem,
[yn+2(𝑥 + 1)2 + nyn+1. 2(𝑥 + 1) +`(n(n-1))/(2!)y_n.2]+`[yn+1. (𝑥 + 1) + nyn (1)] = -4yn
yn+2(𝑥 + 1)2 +(2n + 1)(x + 1) yn+1 + (n2 + 4) yn = 0
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