Consider the equation
$$ \partial_tu =\nabla \cdot [u^{m-1} \nabla u] + \lambda u. $$
I want to consider a transformation that can remove this $\lambda u$ term into a PDE of the form $\partial_tv = D\nabla \cdot [v^{m-1} \nabla v]$ for some constant $D$. Is this possible? I've found that if I let $u = \exp(\lambda t)v$, then
$$ \partial_tv = \exp[\lambda(m-1)t]\nabla\cdot[v^{m-1}\nabla v], $$ but the $\exp[\lambda(m-1)t]$ term remains. Is it possible to remove it with a further transformation of $v$ or $t$?
Was just missing something obvious: Setting $\tau = g(t)$ for some $\tau$ gives $\partial_tv = \partial_\tau v g'(t)$. So, $g'(t) = \exp[\lambda(m-1)t]$ would allow the exponential term to be removed. Thus, $g(t) = 1/[\lambda(m-1)]\exp[\lambda(m-1)t]$ or $t = 1/[\lambda(m-1)]\log[\lambda(m-1)\tau]$.
Also, replacing $\tau$ by $\tau - D/[\lambda(1-m)]$ will make simplify it so that $\tau=0$ when $t=0$.