# Simultaneous DE

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

Originally Posted by daskywalker
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)
• Jan 21st 2010, 12:36 PM
daskywalker
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?
• Jan 21st 2010, 12:41 PM
Jhevon
Quote:

Originally Posted by daskywalker
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