Question

If y=sin[log(x²+2x+1)] then prove that (x+1)²y_n+2 +(2n +1)(x+ 1)y_n+1 + (n²+4)y_n=0.

Answer 1

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,
[y_n+2(𝑥 + 1)² + ny_n+1. 2(𝑥 + 1) +`(n(n-1))/(2!)y_n.2]+`[y_n+1. (𝑥 + 1) + ny_n (1)] = -4y_n

y_n+2(𝑥 + 1)² +(2n + 1)(x + 1) y_n+1 + (n² + 4) y_n = 0