My textbook says the following:
Knowing $y(t)=ce^{rt}$ is a solution for $ay''+by'+cy=0$, we find for the case of $b^2-4ac=0$ that $r=\frac{-b}{2a}$. Therefore one particular solution is $y(t) = \exp\left(\frac{-b}{2a}t \right)$.
However, another linearly independent solution exists and that is $y(t)=t\exp\left(\frac{-b}{2a}t \right)$.
The derivation to find the second equation is that we can say $y(t)=ce^{rt}=v(t)\exp\left(\frac{-b}{2a}t \right)$ with variation of parameters, and then plug that $v(t)\exp\left(\frac{-b}{2a}t \right)$ into the equation (1)
$$ay''+by'+c=0 \tag{1}$$
we find that $v''(t) = 0$ and therefore v(t) is a linear function chosen to be $v(t)=t$. Hence the 2nd solution is $y(t)=te^{rt}$ and solution space is $y(t)=c_1e^{rt}+c_2te^{rt}$.
I have plugged $v(t)\exp\left(\frac{-b}{2a}t \right)$ many times into equation (1) yet can never seem to get the conclusion that $v''(t) = 0$. What does my textbook mean by this?
CITATION: page 213, Differential Equations & Linear Algebra, Second Edition, by Farlow, Hall, McDill, West