Maths Derivatives
posted by Yousef .
If y = x ^ n + ( 1 / x ^ n )
prove that
( x^2y'' ) + ( xy' )  ( n^2y ) = 0

y = x^{n} + x^{n}
y' = nx^{n1}  nx^{n1}
y'' = n(n1)x^{n2} + n(n+1)x^{n2}
x^{2}y'' + xy'  n^{2}y
= n(n1)x^{n} + n(n+1)x^{n} + nx^{n}  nx^{n}  n^{2}x^{n}  n^{2}x^{n}
= 0 
thank u so much
i realised that i had the correct answer but wasn't sure about the signs
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