Asked by Yousef
If y = x ^ n + ( 1 / x ^ n )
prove that
( x^2y'' ) + ( xy' ) - ( n^2y ) = 0
prove that
( x^2y'' ) + ( xy' ) - ( n^2y ) = 0
Answers
Answered by
Steve
y = x<sup>n</sup> + x<sup>-n</sup>
y' = nx<sup>n-1</sup> - nx<sup>-n-1</sup>
y'' = n(n-1)x<sup>n-2</sup> + n(n+1)x<sup>-n-2</sup>
x<sup>2</sup>y'' + xy' - n<sup>2</sup>y
= n(n-1)x<sup>n</sup> + n(n+1)x<sup>-n</sup> + nx<sup>n</sup> - nx<sup>-n</sup> - n<sup>2</sup>x<sup>n</sup> - n<sup>2</sup>x<sup>-n</sup>
= 0
y' = nx<sup>n-1</sup> - nx<sup>-n-1</sup>
y'' = n(n-1)x<sup>n-2</sup> + n(n+1)x<sup>-n-2</sup>
x<sup>2</sup>y'' + xy' - n<sup>2</sup>y
= n(n-1)x<sup>n</sup> + n(n+1)x<sup>-n</sup> + nx<sup>n</sup> - nx<sup>-n</sup> - n<sup>2</sup>x<sup>n</sup> - n<sup>2</sup>x<sup>-n</sup>
= 0
Answered by
Yousef
thank u so much
i realised that i had the correct answer but wasn't sure about the signs
i realised that i had the correct answer but wasn't sure about the signs
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