1. ## Simultaneous DE

dx/dt=-ax

dy/dt=ax-by

where a and b are constants. Solve this system subject to x(0)=x_0 (x "not") and y(0)=y_0 (y "not")

Is it possible to substitute -dx/dt into equation 2 or do I have to solve equation one first to find the solution of x first?

dx/dt=-ax

dy/dt=ax-by

where a and b are constants. Solve this system subject to x(0)=x_0 (x "not") and y(0)=y_0 (y "not")

Is it possible to substitute -dx/dt into equation 2 or do I have to solve equation one first to find the solution of x first?
solve the first DE for x (it is a separable DE), and plug that into the second DE.

(and it is x- and y- naught, as in zero)

3. ok so for the first DE I got x=Ke^-at where K is the constant e^c (from integration) How do I substitute the condition x(0)=x_0? Or do I substitute the whole expression into equation 2?

$\displaystyle x(0) = x_0$ means, when $\displaystyle t = 0,~x = x_0$
We have $\displaystyle x = Ke^{-at}$. applying the initial condition, we see that $\displaystyle x_0 = Ke^0 = K$
Thus, $\displaystyle x = x_0e^{-at}$. Plug this expression in the second equation