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January 27, 2015

January 27, 2015

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

prove that

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

- Maths Derivatives -
**Steve**, Tuesday, October 25, 2011 at 10:12amy = 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 10:25amthank u so much

i realised that i had the correct answer but wasnt sure about the signs

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