Posted by **Yousef** on Tuesday, October 25, 2011 at 8:38am.

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

prove that

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

- Maths Derivatives -
**Steve**, Tuesday, October 25, 2011 at 10:12am
y = x^{n} + x^{-n}

y' = nx^{n-1} - nx^{-n-1}

y'' = n(n-1)x^{n-2} + n(n+1)x^{-n-2}

x^{2}y'' + xy' - n^{2}y

= n(n-1)x^{n} + n(n+1)x^{-n} + nx^{n} - nx^{-n} - n^{2}x^{n} - n^{2}x^{-n}

= 0

- Maths Derivatives -
**Yousef**, Tuesday, October 25, 2011 at 10:25am
thank u so much

i realised that i had the correct answer but wasn't sure about the signs

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