Exponential Solution of Cubic Equation

In summary, the conversation is about finding the solution of a cubic equation in exponential form, which was mentioned in a research article. The equation is (k^6-A k^4+B k^2+E)Y(z)=0 and the given solution is Y(z)=M exp(-k_1z), where M is a constant and k_1 is the solution of the above equation. The person asking the question is having trouble understanding the equation and is looking for help. The respondent suggests using the cubic formula, which can be found through a simple Google search, to solve the equation.
  • #1
adnan jahan
96
0
, Dear Fellows
I need to find the solution of cubic equation in exponential form which is written in a research article

If EQUATION is

(k^6-A k^4+B k^2+E)Y(z)=0

and solution given is as,

Y(z)=M exp(-k_1z)

where M is a constant and k_1 is solution of the above equation.

Any Reply Will Be Informative,
 
Mathematics news on Phys.org
  • #2
I can't make heads or tails out of what you have written. The equation you write:
(k^6-A k^4+B k^2+E)Y(z)=0 simply tells us that, for some z, Y(z)= 0. It does NOT say that this is a cubic equation and Y could be anything.
 
  • #3
adnan jahan said:
, Dear Fellows
I need to find the solution of cubic equation in exponential form which is written in a research article

If EQUATION is

(k^6-A k^4+B k^2+E)Y(z)=0

and solution given is as,

Y(z)=M exp(-k_1z)

where M is a constant and k_1 is solution of the above equation.

Any Reply Will Be Informative,

It looks like you were solving a 6th order differential equation in Y?
[tex][D_z^6 - A D_z^4 + D_z k^2 + E]Y(z)=0[/tex]
There is a cubic formula but it's pretty messy, especially if you are leaving A,B, and E arbitrary. Google "cubic formula" and you'll find many links. The wikipedia article is a good reference: http://en.wikipedia.org/wiki/Cubic_function"
 
Last edited by a moderator:

Related to Exponential Solution of Cubic Equation

1. What is the general form of an exponential solution for a cubic equation?

The general form of an exponential solution for a cubic equation is given by x = Aert, where A is a constant, r is the growth rate, and t is the time variable.

2. How is the growth rate determined for an exponential solution of a cubic equation?

The growth rate for an exponential solution of a cubic equation is determined by finding the complex roots of the equation. The real part of the complex root is the growth rate, while the imaginary part can be used to determine the frequency of the oscillations in the solution.

3. Can an exponential solution of a cubic equation have multiple growth rates?

No, an exponential solution of a cubic equation can only have one growth rate. This is because a cubic equation can have at most three distinct roots, and each root corresponds to a unique growth rate in the exponential solution.

4. How does the initial condition affect the exponential solution of a cubic equation?

The initial condition, or the starting value of the solution at time t = 0, will determine the value of the constant A in the general form of the exponential solution. This means that different initial conditions will result in different exponential solutions, even if they have the same growth rate.

5. Can an exponential solution of a cubic equation ever reach a steady state?

Yes, an exponential solution of a cubic equation can reach a steady state if the growth rate r is negative. In this case, the solution will approach a value of 0 as t approaches infinity, resulting in a stable steady state.

Similar threads

  • General Math
Replies
16
Views
3K
  • General Math
Replies
2
Views
2K
Replies
1
Views
888
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
2
Replies
68
Views
4K
Replies
3
Views
970
Replies
1
Views
2K
  • General Math
Replies
1
Views
4K
  • Calculus and Beyond Homework Help
Replies
7
Views
823
Back
Top