Two differential equations -- Need to find steady state values

In summary, we have found the steady state values for c(t) and k(t) in the given system of equations. By setting c(t) = 0 and solving for k(t), we get k* = 5^4. Plugging this back into the first equation and solving for c(t), we get c* = 0. This means that in steady state, both c(t) and k(t) are equal to 0.
  • #1
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Homework Statement



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I have found the two equations describing the system. They are.

ċ(t) = 1/θ [ f`(k(t)) -(n + δ + β)]c(t)
k̇(t) =f(k(t) - (n +d)k(t) - c(t)
Plugging in the numbers:
ċ(t) = 2 [ 0,25k^(-0,25) -(0,10)]c(t)
k̇(t) =k^0,25 - (0,6)k(t) - c(t)
Since ċ(t) and k̇(t) =0 in steady state ⇒
2 [ 0,25k^(-0,25) -(0,10)]c(t)=0
k̇(t) =k^0,25 - (0,6)k(t) - c(t)=0
So have do I solve this for K* and C*?
I have tried some algebra, but can`t get it to work.

Homework Equations

The Attempt at a Solution

 
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  • #2
To solve for the steady state values, we need to solve the two equations simultaneously.From the first equation, we can see that c(t) must be 0 in order for the equation to be satisfied.Plugging this into the second equation, we get:k̇(t) = k^0,25 - (0,6)k(t) = 0Solving this equation for k(t), we get:k* = 5^4Plugging this back into the first equation, we get:ċ(t) = 2 [ 0,25(5^4)^(-0,25) - (0,10)]c(t) = 0Solving this equation for c(t), we get:c* = 0Therefore, the steady state values are k* = 5^4 and c* = 0.
 

1. What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time, based on the rate at which that quantity is changing.

2. What is a steady state value?

A steady state value is the value that a system settles at over time, where the rate of change is equal to zero. In other words, the system does not change anymore and remains constant.

3. How do you solve for steady state values in differential equations?

To solve for steady state values, you need to set the derivative of the equation equal to zero and solve for the variable. This will give you the value of the variable at steady state.

4. What is the importance of finding steady state values in differential equations?

Steady state values help us understand the long-term behavior of a system and can provide insight into its stability and equilibrium points. They are also useful in applications such as modeling physical systems and predicting future trends.

5. Are there any limitations to using steady state values in differential equations?

Yes, there are limitations to using steady state values in differential equations. These values assume that the system is in equilibrium and does not account for any external disturbances or fluctuations. Additionally, some systems may not have a steady state at all, making it difficult to use this method of analysis.

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