Solve the system of differential equations

In summary: Yes, thank you for the help, I have it solved now for y(t), I think I got it now, the next part of the question which I didn't include initially, asked for: which data (x0,y0) does the solution exist for all time, this is just asking for existence and uniqueness right? Or am I misunderstanding this as...Yes, the solution exists for all time at (x0,y0).
  • #1
ver_mathstats
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21
Homework Statement
Find the solution (x(t),y(t)) to the system.
Relevant Equations
dx/dt = -2x, dy/dt=-y+x^2
I have my set of differential equations which is dx/dt = -2x, dy/dt=-y+x2, with the initial conditions x(0)=x0 and y(0)=y0. I'm a little confused about how to approach this problem.

I thought at first I would differentiate both sides of dx/dt = -2x in order to get d2x/dt2 = -2, and then I would substitute this into the other equation but I don't really know if this is the right approach and I am a little confused. Or would I just solve for x in the first equation and then substitute?

Any help would be appreciated on how to start this problem, thank you.
 
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  • #2
ver_mathstats said:
Homework Statement:: Find the solution (x(t),y(t)) to the system.
Relevant Equations:: dx/dt = -2x, dy/dt=-y+x^2

I have my set of differential equations which is dx/dt = -2x, dy/dt=-y+x2, with the initial conditions x(0)=x0 and y(0)=y0. I'm a little confused about how to approach this problem.

I thought at first I would differentiate both sides of dx/dt = -2x in order to get d2x/dt2 = -2,
Why would you think that would be helpful? Instead, solve the equation dx/dt = -2x to find x(t) as a function of t.

Then replace x2 in the 2nd equation, and you'll have dy/dt + y = <function of t>, which can be solved by straightforward means.
ver_mathstats said:
and then I would substitute this into the other equation but I don't really know if this is the right approach and I am a little confused. Or would I just solve for x in the first equation and then substitute?

Any help would be appreciated on how to start this problem, thank you.
 
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  • #3
Mark44 said:
Why would you think that would be helpful? Instead, solve the equation dx/dt = -2x to find x(t) as a function of t.

Then replace x2 in the 2nd equation, and you'll have dy/dt + y = <function of t>, which can be solved by straightforward means.
Oh thank you, that makes so much sense, I was just over complicating it then. I did solve it to get x(t)=e-2t+C, but we don't worry about the constant till the very end right? Or am I wrong when I say that? So would it just be x(t)=e-2t?
 
  • #4
ver_mathstats said:
Homework Statement:: Find the solution (x(t),y(t)) to the system.
Relevant Equations:: dx/dt = -2x, dy/dt=-y+x^2

I thought at first I would differentiate both sides of dx/dt = -2x in order to get d2x/dt2 = -2, and then I would substitute this into the other equation but I don't really know if this is the right approach and I am a little confused.
No what you saying above is wrong. Differentiating both sides of the aforementioned equation with respect to t, (if we differentiate with respect to x we still don't get what you are saying above) gives instead $$\frac{d^2x}{dt^2}=-2\frac{dx}{dt}$$.
 
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  • #5
Delta2 said:
No what you saying above is wrong. Differentiating both sides of the aforementioned equation with respect to t, (if we differentiate with respect to x we still don't get what you are saying above) gives instead $$\frac{d^2x}{dt^2}=-2\frac{dx}{dt}$$.
Yes I understand that now, I realized I was doing it completely wrong, sorry about that.
 
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  • #6
ver_mathstats said:
Oh thank you, that makes so much sense, I was just over complicating it then. I did solve it to get x(t)=e-2t+C, but we don't worry about the constant till the very end right? Or am I wrong when I say that? So would it just be x(t)=e-2t?
No, you need to put the constant in. ##x(t) = e^{-2t + C} = e^{-2t}e^C## is correct, but you could also write it as ##x(t) =C'e^{-2t}##, where ##C' = e^C##.
 
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  • #7
Mark44 said:
No, you need to put the constant in. ##x(t) = e^{-2t + C} = e^{-2t}e^C## is correct, but you could also write it as ##x(t) =C'e^{-2t}##, where ##C' = e^C##.
Alright thank you, so we just now solve dy/dt+y=Ce-4t using our integrating factor from my understanding?
 
  • #8
ver_mathstats said:
Alright thank you, so we just now solve dy/dt+y=Ce-4t using our integrating factor from my understanding?
Yes, and this one is straightforward. Once you have y(t), you have the solution to the system of equations, because you already have x(t). Finally, you can substitute in the initial conditions x(0) and y(0).
 
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  • #9
Mark44 said:
Yes, and this one is straightforward. Once you have y(t), you have the solution to the system of equations, because you already have x(t). Finally, you can substitute in the initial conditions x(0) and y(0).
Yes, thank you for the help, I have it solved now for y(t), I think I got it now, the next part of the question which I didn't include initially, asked for: which data (x0,y0) does the solution exist for all time, this is just asking for existence and uniqueness right? Or am I misunderstanding this as well?
 
  • #10
Don't forget that ##x\equiv 0## is also a solution. It is not covered by ##e^C##, only by ##C'=0##.
 
  • #11
fresh_42 said:
Don't forget that ##x\equiv 0## is also a solution. It is not covered by ##e^C##, only by ##C'=0##.
Oh okay, thank you, that makes sense.
 

Related to Solve the system of differential equations

1. What is a system of differential equations?

A system of differential equations is a set of equations that describe the relationship between multiple variables and their rates of change. These equations involve derivatives and are used to model various physical, chemical, and biological processes.

2. How do you solve a system of differential equations?

There are various methods for solving a system of differential equations, including separation of variables, substitution, and using matrices. The specific method used depends on the type and complexity of the equations.

3. What is the importance of solving a system of differential equations?

Solving a system of differential equations allows us to understand and make predictions about complex systems in fields such as physics, engineering, and biology. It also helps in developing mathematical models for real-world phenomena.

4. Can a system of differential equations have multiple solutions?

Yes, a system of differential equations can have multiple solutions. This is because the equations may have different initial conditions or parameters, leading to different solutions that can describe the same system.

5. Are there any software or tools available for solving systems of differential equations?

Yes, there are various software and tools available for solving systems of differential equations, such as MATLAB, Mathematica, and Wolfram Alpha. These tools use numerical methods to approximate solutions and can handle complex and large systems of equations.

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