Is there a way to verify the correctness of decoupled linear ODE solutions?

  • Thread starter SteliosVas
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In fact, it would be much better if you wrote the solutions asu(t)= c1e^(2t)+ c2e^(-t)v(t)= c3e^(2t)+ c4e^(-t).Then it is clear that c1 and c3 are coefficients of the "e^(2t)" term and c2 and c4 are coefficients of the "e^(-t)" term.
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SteliosVas
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Okay so i have 2 ODE's i need to decouple them and therefore construct two differential equations each containing just one of the functions... I just wanted to know how can you verify you have the correct solution is more so my question

du/dt= 4u-5v

dv/dt=2u-3v

I end up getting

for u(t) = Ae^x + Be^-3x

and for v(t) = Ae^5x +Be^x

My question is also when assigning the solutions to the characteristic equation do you give the higher number to a and the lower to b??

THanks :)
 
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  • #2
You can check your solutions by substituting them back into the original problem. For example, if you calculate ## du/dt ## using your expression for ## u ##, then that should be equal to ## 4u-5v ## using both your expressions for ## u## and for ## v ##. I think you will find that your solution is not correct.
 
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  • #3
SteliosVas said:
Okay so i have 2 ODE's i need to decouple them and therefore construct two differential equations each containing just one of the functions... I just wanted to know how can you verify you have the correct solution is more so my question

du/dt= 4u-5v

dv/dt=2u-3v

I end up getting

for u(t) = Ae^x + Be^-3x

and for v(t) = Ae^5x +Be^x

My question is also when assigning the solutions to the characteristic equation do you give the higher number to a and the lower to b??
As already stated by Geofleur, your solutions don't work. Did you solve this system by matrix diagonalization? If so, please show us your work in getting the eigenvalues.
 
  • #4
As Geofleur and Mark44 say, you do not have the correct solutions to the equations. You should have arrived at a characteristic equation [itex]r^2- r- 2= 0[/itex].

But to answer your question, "A" and "B" are both unknown constant coefficients. It doesn't matter which you assign to which exponential.
 

FAQ: Is there a way to verify the correctness of decoupled linear ODE solutions?

1. What is a decoupled linear ODE?

A decoupled linear ODE (Ordinary Differential Equation) is a type of differential equation where the dependent variables are not explicitly dependent on each other, and the coefficients are constant. This means that the equation can be separated into individual equations for each variable, making it easier to solve.

2. How do you decouple a linear ODE?

To decouple a linear ODE, you can use a technique called substitution. This involves replacing one of the variables with a new variable, which eliminates it from the equation. Then, you can solve the new equation for the remaining variable. This process can be repeated for each variable, resulting in a set of decoupled equations.

3. What are the benefits of decoupling linear ODEs?

Decoupling linear ODEs can make them easier to solve, as each equation can be solved separately. It also allows for a better understanding of the relationship between the variables, as they are no longer dependent on each other. This can be especially useful in modeling real-world systems.

4. Are there any limitations to decoupling linear ODEs?

Decoupling linear ODEs is not always possible, and it may not always result in a simpler form of the equation. It also requires the coefficients to be constant, which may not always be the case. Additionally, the process of decoupling can be time-consuming and may not be worth the effort for simpler equations.

5. How are decoupled linear ODEs used in scientific research?

Decoupled linear ODEs are commonly used in mathematical modeling to describe the behavior of complex systems in physics, chemistry, biology, and other fields. They can also be used to analyze and predict the behavior of systems over time, making them a valuable tool in scientific research.

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