Exponential Solution of Cubic Equation

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The discussion centers on finding the exponential solution of a cubic equation presented in a research article, specifically the equation (k^6 - A k^4 + B k^2 + E)Y(z) = 0. The proposed solution is Y(z) = M exp(-k_1z), where M is a constant and k_1 is derived from the equation. Participants clarify that the equation represents a sixth-order differential equation rather than a cubic equation, which complicates the solution process. Acknowledgment is made that while there is a cubic formula, it can be complex, especially with arbitrary coefficients A, B, and E. The discussion suggests looking up the cubic formula for further insights.
adnan jahan
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, 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,
 
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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.
 
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?
[D_z^6 - A D_z^4 + D_z k^2 + E]Y(z)=0
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"
 
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