# Simultaneous DE

• January 21st 2010, 12:20 PM
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?
• January 21st 2010, 12:25 PM
Jhevon
Quote:

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)
• January 21st 2010, 12:36 PM
$x(0) = x_0$ means, when $t = 0,~x = x_0$
We have $x = Ke^{-at}$. applying the initial condition, we see that $x_0 = Ke^0 = K$
Thus, $x = x_0e^{-at}$. Plug this expression in the second equation