What you've done looks perfectly fine.
Here's a general outline of what you do when you're looking to find where solutions exist for the first-order differential equation $y'+ p(t) y = g(t)$.
Write the differential equation in the form: $y'=f(y, t)$
Find $f_y = \frac{\partial}{\partial y} f$.
Determine points of discontinuities of both $f_y$ and $f$.
At this point, if you're just looking to see if a particular initial condition ($t_0$) has a solution, just check if $t_0$ is one of the points of discontinuity.
If you're looking for where the solution exists:
- Draw a number line denoting where the discontinuities are (if possible).
- Find where the initial condition falls on the number line.
- If the discontinuity to the left of $t_0$ is $a$, and the discontinuity to the right of $t_0$ is $b$, then the solution exists on the interval $(a, b)$.
EDIT
Based on requests from comments below, here's a statement of the existence and uniqueness theorem:
Let the functions $f$ and $\frac{\partial f}{\partial y}$ be continuous in some rectangle $\alpha < t < \beta$, $\gamma < y < \delta$ containing the point $(t_0, y_0)$. Then, in some interval $t_0 - h < t < t_0 + h$ contained in $\alpha < t < \beta$, there is a unique solution $y = \phi(t)$ of the initial value problem:
$$\begin{array}{cc}
y' = f(t,y) & y(t_0) = y_0.
\end{array}$$
Source: Elementary Differential Equations and Boundary Value Problems, Boyce and DiPrima, 10th Edition, pg 70.