I'm given the initial value problem:
$x' = \frac{3}{t}x + e^{3t}$, $x(1) = 2$
Using the variation of parameters formula, I end up with:
$2e^{3 \ln|t|} + e^{3 \ln|t|} \int^{t}_{1}e^{-3\ln|s|} e^{3s} ds$
which simplifies to:
$2|t|^{3} + |t|^{3} \int^{t}_{1} |s|^{-3} e^{3s} ds$
How can I justify ignoring these absolute values? The only justification I see is that the integral is from $1$ to $t$, implying that $t$ is positive (but not really). However, this is just from following the variation of parameters formula - where does this assumption that $t$ is positive actually originate from in the beginning?
The equation is not defined for $t=0$. So the interval where the solution is defined must be contained in $(-\infty,0)$ or in $(0,+\infty)$. Since the initial time is $t=1$, you need to consider an interval in $(0,+\infty)$.