Question

If y=acos(log x)+bsin(log x) then prove that x^2yn+2 + (2n+1)xyn+1 + (n^2+1)yn = 0

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

We have y = acos(log x) + b sin (log x)

⇒ y1 = – asin(log x).1/x + bcos(log x).1/x

⇒ xy1 = –asin(log x) + b cos(log x)

⇒ xy2 + y1 = –a cos(logx).  1/x – bsin (logx).1/x

⇒ x2y2 + x y1 = –y ⇒ x2y2 + xy1 + y = 0

By Leibnitz theorem, we get,

x2 yn+2 + 2(n + 1) x yn+1 + n(n – 1) yn + x yn+1 + nyn + yn = 0

⇒ x2 . yn+2 + (2n + 1)x . yn+1 + (n2 + 1)yn = 0