Question

show that y=x^2-x^(-1) is a solution of the differential equation y"-2y/x^2=0

Answer 1

If \[y = x ^{2} - x ^{-1}\], then \[y' = 2x + x ^{-2}\], and \[y'' = 2 - 2x ^{-3}\]. Given the original value of y, \[2y/x ^{2} = (2x ^{2} - 2x ^{-1})/x ^{2} = 2 - 2x ^{-3} = y''\]

Answer 2

That's very helpful, merci beaucoup!