Solution to an ODE Let $g(t,z)$ be a continuous complex-valued function. Here, $t\in[-T,T]$ for some positive real  number $T$ and $z$ is complex one-dimensional. Denote $\partial / \partial t$ by $D_t$. How do we solve a differential equation of the form
$$(tD_t + \lambda)u = g$$
where $\lambda$ is a complex constant? Feel free to introduce additional assumptions if needed.
Additional Info: I have read that the solution is
$$u(t,z) = \int_{[0,1]} s^{\lambda - 1} g(st,z) ds. $$
 A: As Guido suggested, you can just think of this as single-variable ODE:
$$ \begin{align}
\left(t \frac{\partial}{\partial t} + \lambda\right) u(t, z) &= g(t, z) \\
\frac{\partial u}{\partial t}(t, z) + \frac{\lambda}{t} u(t, z) &= \frac{1}{t} g(t, z).
\end{align} $$
The standard procedure here is to evaluate the integrating factor, [*]
$$ \exp \int_1^{t} \frac{\lambda}{\tau} \operatorname{d}\! \tau = t^\lambda $$
and then proceed to solve the differential equation after multiplying by the integrating factor:
$$ \begin{align}
t^\lambda \frac{\partial u}{\partial t}(t, z) + \lambda t^{\lambda - 1} u(t, z) &= t^{\lambda - 1} g(t, z) \\
\frac{\partial}{\partial t} t^\lambda u(t, z) &= t^{\lambda - 1} g(t, z) \\
u(t, z) &= \frac{1}{t^\lambda} \int_C^t \tau^{\lambda - 1} g(\tau, z) \operatorname{d}\! \tau.
\end{align} $$
Now, the solution mentioned in the question is indeed correct as well, aside from the sign error on $\lambda$ (it's easy to verify it).  Thanks to automaton3's hint, I realized that it's actually just a simple matter of rescaling the integration variable.  Let $\tau = s t$ and assume $C = 0$,
$$ \begin{align}
u(t, z) &= \frac{1}{t^\lambda} \int_0^t (s t)^{\lambda - 1} g(s t, z) \operatorname{d} (s t) \\
&= \frac{1}{t^\lambda} \int_0^t (s t)^{\lambda - 1} g(s t, z) \operatorname{d} (s t) \\
&= \frac{1}{t^\lambda} \int_{0 / t}^{t / t} t^\lambda s^{\lambda - 1} g(s t, z) \operatorname{d}\! s
\\
&= \int_0^1 s^{\lambda - 1} g(s t, z) \operatorname{d}\! s.
\end{align} $$
[*] For simplicity, the lower bound of integration were chosen arbitrarily.
