I have the series
$$T(x) = \sum_{n=0}^\infty t_nx^n$$
Is there anyway to express $\sum_{n=0}^\infty nt_nx^n$ in terms of $T(x)$? I tried differentiating $T(x)$ but only got so far as to express the latter as $xT'(x)$.
If that desired form of $T(x)$ is not possible, then is there a way to solve the following for $T(x)$? $$x^2T'(x) +(3x-1)T(x) +1 = 0$$ My goal is to obtain an algebraic expression for $T(x)$, preferably one that could be then manipulated to obtain an infinite power series.
You are correct: $$\sum_{n=0}^\infty n t_n x^n = x T'(x)$$.
$$ x^2 T'(x) + (3x-1) T(x) + 1 = 0$$ is a first-order linear differential equation. Its general solution is $$ c \frac{\exp(-1/x)}{x^3} - \frac{\exp(-1/x) \Gamma(0,-1/x)}{2 x^3} - \frac{1}{2x} - \frac{1}{2x^3} $$ where $c$ is an arbitrary constant. However, this is not analytic at $x=0$.
EDIT: In terms of the $t_n$, your differential equation says $t_0 = 1$ and $t_n = (n+2) t_{n-1}$ for $n \ge 1$. This recurrence can easily be solved, to obtain $t_n = (n+2)!/2$. However, the series $\sum_{n=0}^\infty (n+2)!\; x^n/2$ diverges for all $x \ne 0$.