# Math Help - Reduction of Order

1. ## Reduction of Order

Hello,

I have the equation $t^2y''-2y=0$, where t>0, and I know that
$y_1(t)=t^2$.

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, $y_1(t)=t^2$. I then found the derivatives of all the $y_2$'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.

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

Plugging in: $t^2(t^2v''+2vt'+4tv')-2(t^2v)=0
=> t^4v''+2t^2vt'+4t^3v'-2t^2v=0$

And this leads me to believe that the terms $2t^2vt'$ and $-2t^2v$ 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!

2. The DE is of the 'Euler's type' and its solutions are of the form $y= t^{\alpha}$... imposing that You find a second order algebraic equation in $\alpha$ one solution of which is $\alpha=2$... what is the other solution?...

Kind regards

$\chi$ $\sigma$

3. Hi!

Thanks for the quick response, but unfortunately your hint escapes me, as the 'other solution' is exactly what I'm trying to find through reduction of order - which I can't seem to do correctly, hence my problem.

4. Originally Posted by endreoc
Hello,

I have the equation $t^2y''-2y=0$, where t>0, and I know that
$y_1(t)=t^2$.

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, $y_1(t)=t^2$. I then found the derivatives of all the $y_2$'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.

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

Plugging in: $t^2(t^2v''+2vt'+4tv')-2(t^2v)=0
=> t^4v''+2t^2vt'+4t^3v'-2t^2v=0$

And this leads me to believe that the terms $2t^2vt'$ and $-2t^2v$ 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!
I get

$y_{2}(t)=t^{2}v\implies y_{2}'(t)=2tv+t^{2}v'\implies
y_{2}''(t)=2(v+tv')+2tv'+t^{2}v''$

$=2v+4tv'+t^{2}v''.$

You'll notice a few differences between this result and yours. In particular, there is no $t',$ because I'm taking the derivative with respect to $t.$

Plugging into the DE yields

$t^{2}(2v+4tv'+t^{2}v'')-2t^{2}v=0\implies 4t^{3}v'+t^{4}v''=0.$

And now you can reduce the order, cancel some $t$'s, and proceed on your merry way. Alternatively, you could notice that, right now, the LHS of the above equation is a perfect derivative, allowing you to simply integrate almost immediately. That is, you have

$(t^{4}v')'=0.$