Removing transients in highpass filtering with MATLAB
Your first order filter recursion for some real constants $a,b,c$ is 

$$ y[n] = a x[n] + b x[n-1] - c y[n-1] $$

with the two initial memory states $x[-1]$ and $y[-1]$ at $n=0$. 

Your "no transient" condition can be translated to $y[0]=0$ and a necessary second condition so that you can solve for both of your memory states. The second condition could be, that the discrete derivative of $y$ also vanishes at $n=0$, so $y[0]-y[-1]=0$. You can also take any other condition that seems sensible to you.

The two equations give you a unique solution for the two unknown memories, namely:
$$y[-1] = 0$$
 and 
$$x[-1] = -\frac{a}{b}x[0] $$

Alternatively, your conditions may better be chosen as $y[0]=0$ and $y[0]-y[-1]=x[0]-x[-1]$ in order to capture the initial slope of the input. The resulting recursion equation at $n=0$ is then
$$0=a x[0]+b x[-1]+c(x[0]-x[-1])$$
giving you the solution
$$x[-1]=-\frac{a+c}{b-c}x[0]$$
and
$$ y[-1]= -\left(1+\frac{a+c}{b-c}\right)x[0]$$

(Please check my calculations!)

But in general you cannot expect a simple initial condition to give you the same result as knowing the signal history. So you can only take this to a certain point and in general it would probably be better if you just discarded the transient response of your output.