Originally Posted by

**endreoc** Hello,

I have the equation $\displaystyle t^2y''-2y=0$, where t>0, and I know that

$\displaystyle y_1(t)=t^2$.

I would then like to use reduction of order to find some v(t), but I am getting a bit stuck, so I will just show my workings and hopefully someone will pick up on what I am doing wrong, as I remain with not only the derivatives of v(t) but v(t) itself, and am unsure of how to remove it so I can proceed.

v(t) is the function I'd like to find with my current known solution, $\displaystyle y_1(t)=t^2$. I then found the derivatives of all the $\displaystyle y_2$'s, plugged in, but I could not proceed as I still had the v(t) term whereas I'm supposed to be only left with its derivatives.

$\displaystyle y_2(t)=t^2v, y_2'(t)=2tv+t^2v'', y_2''(t)=2vt'+2tv'+2tv'+t^2v''$

Plugging in: $\displaystyle t^2(t^2v''+2vt'+4tv')-2(t^2v)=0

=> t^4v''+2t^2vt'+4t^3v'-2t^2v=0$

And this leads me to believe that the terms $\displaystyle 2t^2vt'$ and $\displaystyle -2t^2v$ are the ones that are 'supposed' to cancel out and that I'm adding one too many t' to the equation, but I don't really see how that's wrong 'cause it's what I get when I repeat it, as well.

Hopefully someone will spot it, thanks!