How is this concluded in the following differential equation? Question: Solve the differential equation $$t^2x''-2tx'+(t+2)x=t^6$$
Answer:$$x(t)=u(t)v(t); x'=u'v+uv' ; x''=u''v+2u'v'+uv'';$$
       $$t^2(u''v+2u'v'+uv'')-2t(u'v+uv')+(t^2+2)uv= t^6 $$ then: 
       $$ t^2uv''+ \underbrace{(2t^2u'-2tu)}_{=0}v'+\underbrace{(t^2 u''-2tu'+(t^2+2)u)}_{=0}v=t^6  |:t^3$$
from here the the following is unclear to me:
$$v''+v=t^3.$$ the answer goes on from here being solved as a homegeous equation, quite simply.. Also what is unclear is how $\underbrace{(2t^2u'-2tu)}_{=0} $and $\underbrace{(t^2 u''-2tu'+(t^2+2)u)}_{=0}$. The motivation behind this technique piques my interest..

By making a trial factorization $y=u·v$ you give yourself a degree of freedom that you can use as you like. Compare with the solution of cubic equations.
Here you use $u$ to impose the condition that the coefficient of $v'$ vanishes. One finds that the solution is $u=t$. Inserting into the equation gives $t^3v''+t^3v=t^6$, the coefficient of $v$ is not zero, the corresponding subscript is wrong.
