1. The problem statement, all variables and given/known data
Hi, i was reviewing some of the material covered this week and i'm wondering if someone can explain a brief step that i'm not quite understanding. The step is once you differentiate your given solution for example twice, then plug it back into the original equation, one of the terms becomes zero. I can't figure out why this is happening if someone can explain that would be apprieated.

3. The attempt at a solution

Take this example:

x^{2}y'' + 2xy' - 6y=0
the given solution is y1= x^{2} and we must find y2.[/quote]
Well, the fundamental problem is that [itex]x^2[/itex] is NOT a solution to this equation!

If y= x^2, then y'= 2x and y''= 2 so putting that into the equation 2+ 4x^2- 6x^2= 2- 2x^2 which is NOT 0.

Check the equation again. Are you sure it was not [itex]x^2y''+ 2xy'- 6x^2y= 0[/itex]?