It is not necessary to solve the DE to answer this question.
Suppose that $y(t)$ is a solution to the DE that touches, but does not cross, the $t$-axis, and let $t_0$ denote the time at which this happens.
Since the solution $y$ touches the $t$-axis at the time $t_0$, we have $y(t_0) = 0$.
Since the solution $y$ does not cross the $t$-axis at this time, we must have $y'(t_0) = 0$. (If $y'(t_0)$ were positive, then there would be some $\delta > 0$ with the property that $y(t)$ is negative for all $t$ in $(t_0 - \delta, t_0)$ and positive for all $t$ in $(t_0, t_0 + \delta)$, and hence the solution $y$ crosses the $t$-axis instead of just touching touch it. The analysis is similar if $y'(t_0) < 0$.)
Since $y$ solves the DE, we have
y'(t_0) + 5 y(t_0) = 5 t_0.
Using the facts established earlier that $y'(t_0) = y(t_0) = 0$, we deduce from the above equation that $0 = 5 t_0$, and hence that $t_0 = 0$. It follows that $$
y(0) = y(t_0) = 0.
[To make the argument in "if $y'(t_0)$ were positive..." more rigorous, you need to observe that if $y$ solves the DE, then $y$ is differentiable, and hence continuous, and hence $y' = 5t - 5y$ is continuous. So the continuous function $y'$, being positive at $t_0$, must also be positive in a neighborhood of $t_0$. The claims about the sign of $y$ then follow from the mean value theorem.]