Solving IVP for Lotka Volterra equations Let
$x'=x(a-by) \\ y'=y(-c+dx)$
$a,b,c,d \in \mathbb R^+$
I want to determine the solution of the IVP $(x_0,0)$
I tried:
$y'(t_0)=0 \Rightarrow y(t)=y'(t)=0 \ \forall t$.
Therefore $x'(t_0)=x(t_0)(a-by(t_0))=x(t_0)a$.
Solving this leads to
$ x(t_0)=\tilde ce^{at_0}\overset{!}{=}x_0$
$\Rightarrow \tilde c=x_0e^{-at_0}$
And therefore $x(t)=\tilde ce^{-at}=x_0e^{a(t-t_0)}$
I'm not sure if the first implication, namely $y'(t_0)=0 \Rightarrow y(t)=y'(t)=0 \ \forall t$ is correct. Is it?
And I also saw $x(t)=x_0e^{at}$ as a solution online, so where are my mistakes?
 A: You wrote :

$y'(t_0)=0 \Rightarrow y(t)=y'(t)=0 \ \forall t$.

This is not correct as shown below.
$\frac{dx}{dt}=x(t)(A-By(t)) \\ \frac{dy}{dt}=y(t)(-C+Dx(t))$ ,
For a given $t_0$ which is a constant, not a variable, we have $y(t_0)=0$ and then :
$y_{(t=t_0)}=\left(\frac{dy(t)}{dt}\right)_{(t=t_0)}=0$ .
$y_{(t=t_0)}$ is not a function of $t$ . So one cannot differentiate with respect to $t$. Thus $y'(t)$ is meaningless or $y'(t)=0$ and $y(t)=$constant which is not correct. 
Of course one can consider the whole family of functions $y(t)$ for various $t_0$. This is no longer a function of one variable $y(t)$, but this is a function of two variables $y(t_0,t)$.
Then one can differentiate with respect to $t_0$ or with respect to $t$. But one cannot confuse $\frac{\partial y}{\partial t_0}$ with $\frac{\partial y}{\partial t}$.
Thus $y'(t_0)\neq y'(t) \ \forall t$ . The symbolism with apostrophe is confusing. This is probably the cause of the mistake.
$y'(t_0)=0$ doesn't imply $y(t)=y'(t)=0 \ \forall t$.
By the way, if $y(t)=y'(t)=0 \ \forall t$ was correct, this would imply $y(t)=$constant, which is obviously false except for the trivial solution $y(t)=0$ .
