If y = x ^ n + ( 1 / x ^ n )

Question

prove that

( x^2y'' ) + ( xy' ) - ( n^2y ) = 0

Answer 1

y = xⁿ + x^-n
y' = nx^n-1 - nx^-n-1
y'' = n(n-1)x^n-2 + n(n+1)x^-n-2

x²y'' + xy' - n²y
= n(n-1)xⁿ + n(n+1)x^-n + nxⁿ - nx^-n - n²xⁿ - n²x^-n
= 0

Answer 2

thank u so much
i realised that i had the correct answer but wasn't sure about the signs