Question: First-order differential equations: the difference between forward and backward solution

Hello all, 

I have the system which consists of three linear first-order differential equations (see below). First, I solved this system numerically backward in time by implying the final condition for the variables at time t=T (i.e, x(T=4)=10 and so on) and depicted the graph. Then, from the obtained solution I found the values of the variables at time t=0 and solve once again the system of ODE's but forward in time with earlier obtained initial conditions for the variables at time t=0. I presumed that I should receive the same plot (solution) but I didn't. I would like to ask if there is any mistake or the solution of the ODE's solved forward or backward in time should be quite different?      

sys := {diff(u(t), t) = 4*u(t), diff(x(t), t) = 6*x(t)+2, diff(z(t), t) = x(t)+u(t)}

Thanks in advance,