Differential Equations - First Order Systems

In summary: You're fine. I appreciate the help. Have a good one.In summary, the problem involves solving a system of differential equations where dx/dt = 3x + y and dy/dt = -y. The solution involves separating dy/dt and integrating to find y = Ce^-t. This is then plugged into the dx/dt equation and solved using the Integrating Factor method to get x = (-C/3) + Ce^(3t). However, there was a mistake in the integration of e^(-4t) which led to an incorrect solution. After correcting the integration, the correct solution is x = -C/4 e^(-t) + K e^(3t).
  • #1
twiztidmxcn
43
0
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
 
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  • #2
twiztidmxcn said:
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.
so far so good
x = (-1/3)C + Ce^(3t)
I don't get this. I get [itex] x(t) = - \frac{1}{4C} e^{-Ct} [/itex].
You have done a mistake either in integrating or in solving for x.
 
  • #3
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
twiztidmxcn said:
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
Sorry, I had made a few mistakes .

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

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

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

[itex] x = -\frac{C}{4} e^{-t} + K e^{3t} [/itex]
 
Last edited:
  • #5
twiztidmxcn said:
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.
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
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:
  • #7
twiztidmxcn said:
Haha, I just caught my error. Ima give it a whirl now with the CORRECT integral, hehe.

Thank you nrged

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

Best luck!
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It involves one or more variables and their rates of change, and is commonly used to model real-world phenomena in fields such as physics, engineering, and biology.

2. What is a first order system?

A first order system is a system of differential equations that involves only first derivatives. This means that the highest order of derivative in the equation is one. First order systems are commonly used to model simple systems that have a single input and single output, such as the motion of a particle under the influence of a force.

3. How do you solve a first order system of differential equations?

First order systems can be solved using a variety of methods, including separation of variables, substitution, and integration. The specific method used will depend on the form of the equations and the initial conditions given. It is important to check for solutions that may not be valid in certain regions of the domain.

4. What are the applications of first order systems?

First order systems are used to model a wide range of phenomena in fields such as physics, engineering, and biology. They can be used to describe the motion of particles, the growth of populations, the flow of fluids, and many other real-world processes. They are also used in control theory and signal processing to design systems that can respond to different inputs.

5. What are some challenges in solving first order systems?

One of the main challenges in solving first order systems is finding an analytical solution. In many cases, it may not be possible to find an exact solution, and numerical methods must be used instead. Another challenge is dealing with complex systems that involve multiple variables and nonlinear relationships, which can make it difficult to find a solution that accurately models the system's behavior.

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