1. ## Reduction of Order

$\displaystyle t^2y'' - 4ty' + 6y$

$\displaystyle y1(t) = t^2$
given one solution find another solution. I solved this using reduction of order.

I got

$\displaystyle v''t^4 = 0$

where do I proceed from here tho?

I know I have to integrate it or something but im forgetting the steps.

2. Originally Posted by p00ndawg
$\displaystyle t^2y'' - 4ty' + 6y$

$\displaystyle y1(t) = t^2$
given one solution find another solution. I solved this using reduction of order.

I got

$\displaystyle v''t^4 = 0$

where do I proceed from here tho?

I know I have to integrate it or something but im forgetting the steps.
I assume the above equaiton is equal to zero....

So the 2nd solution is of the form...

$\displaystyle y=t^2\cdot u(t)$

$\displaystyle y'=2tu+t^2u'$

$\displaystyle y''=2u+2tu'+2tu'+t^2u''=t^2u''+4tu'+2u$

Plugging in we get

$\displaystyle t^2(t^2u''+4tu'+2u)-4t(2tu+t^2u')+6(t^2u)=0$

$\displaystyle t^4u''+4t^3u'+2t^2u-8t^2u-4t^3u'+6(t^2u)=0$

$\displaystyle t^4u''=0$

So this implies that $\displaystyle u''=0$

Now integrating twice we get

$\displaystyle u'=c$ and $\displaystyle u=ct+d$ so if we choose c=1 and d=0 we get

$\displaystyle u=t$

so the 2nd solution is

$\displaystyle y_2=c_2t^3$