# Differential Equations - First Order Systems

1. Jun 27, 2007

### twiztidmxcn

Solve:

dx / dt = 3*x + y

dy / dt = -y

As for solving this, here is what I've got so far:

Since dy/dt is separable, I found that dy / y = -dt, integrated, and solved explicitly for y:

y = Ce^-t

I then plugged Ce^-t for y in the dx / dt portion of the system and found that:

dx/dt - (3*x) = Ce^-t

Then used Integrating Factor method, found u(t) = e^(-3t), multiplied to both sides and integrated and solved explicitly for x.

x = (-1/3)C + Ce^(3t)

For the second portion of the question, it asks to verify that the solution is correct.

To do that, I plugged in the x and y solutions from part 1, as well as took derivatives of x and y and plugged them in as well.

dy / dt = -y, -Ce^-t = -Ce^-t

However, the problem I am having is that when I plug in for the dx / dt equation, I get down to t = ln1 = 0.

I don't understand if this is wrong or what, as the y solution works but the x one doesn't seem to come to any reasonable conclusion.

Any help in understanding these results, or in finding mistakes would be much appreciated.

-twiztidmxcn

2. Jun 27, 2007

### nrqed

so far so good
I don't get this. I get $x(t) = - \frac{1}{4C} e^{-Ct}$.
You have done a mistake either in integrating or in solving for x.

3. Jun 27, 2007

### twiztidmxcn

hrm, I've re-done the integration multiple times and keep getting the same thing.

I am integrating this:

(x*e^(-3t))' = C*e^(-4t)

After integrating I get:

x*e^(-3t) = ( -C / [3e^(3t)] ) + C

Divide out the u(t) = e^(-3t) and end up with this:

x = (-C / 3) + C*e^(3t)

I've also done this integration in both Maple and Mathematica and got the same thing as what I got by hand.

So...I'm not sure its the integration or solving for x, I'm baffled.

4. Jun 27, 2007

### nrqed

I did forget the constant of integration (sorry) and I made typos by putting the C in the denominator instead of the numerator and putting it in the exponential (it's getting too late for me to do simple maths!).
here is my work

$d/dt(x e^{-3t}) = C e^{-4t} \rightarrow x e^{-3t} = -\frac{C}{4} e^{-4t} + K$

wher I use K for the constant of integration to distinguish it from C. Solving, we get

$x = -\frac{C}{4} e^{-t} + K e^{3t}$

Last edited: Jun 27, 2007
5. Jun 27, 2007

### nrqed

It's your integration of e^(-4t) which is incorrect.
You may check that my answer satisfies the DE for x(t)

regards

Patrick

6. Jun 27, 2007

### twiztidmxcn

Haha, I just caught my error. Ima give it a whirl now with the CORRECT integral, hehe.

Thank you nrged

*edit* - got it, much thanks to you nrged

Last edited: Jun 27, 2007
7. Jun 27, 2007

### nrqed

Good! And you are welcome. Sorry for my mistakes.

Best luck!