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- #1

So I have the equation

$(x-1)y''-xy+y=0$

with x being a solution. Then:

$y(x)=xv(t)$

$y'(x)=v(t)+xv'(x)$

$y''(x)=2v'(x)+xv''(x)$

I dont see where the 2 is coming from in the last line as my previous understanding of it was that:

$y''=x''v(x)+x'v'(x)+xv(x)$

so I thought it would just be one v'(x) with the v(x) disappearing?

Thanks

Carla