How do I solve these coupled Differential Equation?

In summary, the conversation discusses determining the order of coupled differential equations, the best methods for solving them, checking the validity of solutions, limitations of solving them, and the importance of carefully analyzing equations to choose the appropriate method. It is mentioned that the order of a coupled differential equation is determined by the highest derivative present and that there are multiple methods for solving them, such as separation of variables and numerical methods. It is also noted that solutions can be checked by plugging them back into the original equations or using software/graphing techniques. However, solving coupled differential equations can be challenging due to their complexity and the potential for inaccurate solutions.
  • #1
baouba
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Homework Statement


dNa/dt = -Na/Ta where Na is the function and Ta is the constant
dNb/dt = Na/Ta - Nb/Tb where Nb is the function and Tb is the constant

Homework Equations


My Prof said Nb(t) has the form Nb(t) = Cexp(-t/Ta) + Dexp(-t/Tb)

The Attempt at a Solution


I know the first equation solves to Na(t) = Na(0)exp(-t/Ta)

The second equation can be written,

dNb/dt = (TbNa-TaNb)/(TaTb)

(TaTb)dNb/dt = TbNa-TaNb

Separation of variables and integrating gives:

(TaTb) ∫ [TbNa-TaNb]^-1 dNb = ∫dt

(TaTb)(-1/Ta)ln(NaTb - NbTa) = t + C

Rearranging,

Nb = (Tb/Ta)Na - Cexp(-t/Tb)

Subbing in Na,

Nb = (Tb/Ta)(Na(0)exp(-t/Ta)) - Cexp(-t/Tb)

at t = 0,

Nb(0) = (Tb/Ta)(Na(0)) - C

so C = (Tb/Ta)(Na(0)) - Nb(0)

Subbing back into Nb(t),

Nb = [(Tb/Ta)(Na(0)exp(-t/Ta))] - [(Tb/Ta)(Na(0)) - Nb(0)]exp(-t/Tb)

Apparently the right answer is,

Nb = (1 / ((Ta/Tb)-1.)) Na(0)exp(-t/Ta)+ (Nb(0)-(Na(0)/((Ta/Tb)-1.)))exp(-t/Tb);

but I just can't seem to get it. Can anyone tell me where I went wrong?

Thanks
 
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  • #2
baouba said:

Homework Statement


dNa/dt = -Na/Ta where Na is the function and Ta is the constant
dNb/dt = Na/Ta - Nb/Tb where Nb is the function and Tb is the constant

Homework Equations


My Prof said Nb(t) has the form Nb(t) = Cexp(-t/Ta) + Dexp(-t/Tb)

The Attempt at a Solution


I know the first equation solves to Na(t) = Na(0)exp(-t/Ta)

.........

Thanks

Once you have solved the first equation for ##N_a(t)##, your second equation becoms
[tex] \frac{dN_b}{dt} = f(t) - \tau_b N_b, [/tex]
where ##\tau_b \equiv 1 / T_b## is a constant and ##f(t) = N_a(t)/T_a## is a known function. You can solve this DE using an integrating-factor approach; see, eg.,
https://en.wikipedia.org/wiki/Integrating_factor .
 

1. How do I determine the order of the coupled differential equations?

The order of a coupled differential equation is determined by the highest derivative present in the equation. For example, if the highest derivative is a second derivative, the equation is considered a second-order coupled differential equation.

2. What is the best method for solving coupled differential equations?

The best method for solving coupled differential equations depends on the specific equations and initial conditions. Some common methods include separation of variables, substitution, and numerical methods such as Euler's method or Runge-Kutta methods.

3. How do I check if my solution to a coupled differential equation is correct?

You can check the validity of your solution by plugging it back into the original equations and seeing if it satisfies all of the equations. You can also use computer software or graphing techniques to visualize and confirm the solution.

4. Can I use the same method to solve all coupled differential equations?

No, different types of coupled differential equations may require different methods for solving. It is important to carefully analyze the equations and choose the appropriate method for each problem.

5. What are the limitations of solving coupled differential equations?

The complexity of coupled differential equations can make them difficult to solve analytically, and numerical methods may not always provide accurate solutions. Additionally, the initial conditions and parameters of the equations may affect the stability and validity of the solutions.

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