Cubic equations with independent variables

[omarxx84]

Hi colleages. can you help me to solve the cubic equation below:
2N(Ep-En)hp^3(x)-3[M(x,t)(Ep-En)-2NEnh]hp^2(x)-6Enh[M(x,t)+Nh]hp(x)+Enh^2[3M(x,t)+2Nh]=0 notice that all variables in the equation are dependent on x only, except M is dependent on x and t.
En, Ep, N and h are constants.
i know the solution when the equation dependent on one variable, but the problem is that M dependent on two variables x and t.
please help me and i will be very grateful for you...

 

[Mark44]

Please don't double post! (Or triple post!) This is the third thread that you have started with exactly the same problem: https://www.physicsforums.com/showthread.php?t=407426 and https://www.physicsforums.com/showthread.php?t=403107.

In one of the other threads someone asked you to rewrite the equation in a less dense form. With everything so tightly packed together, the equation is difficult to comprehend.

 

[omarxx84]

i am so sorry,

 

[Mark44]

There is an algorithm for finding the roots of a cubic equation - see http://en.wikipedia.org/wiki/Cubic_equation. Using this algorithm might help you solve for the values of p(x) that are roots of your equation.

You haven't said anything about p(x). What information do you have about this function?

 

[CellCoree]

Hello,

Thank you for your question. Solving cubic equations with independent variables can be a challenging task, but I will do my best to help you.

First, let's rewrite the given equation in a more simplified form:

2N(Ep-En)hp^3(x) - 3[M(x,t)(Ep-En) - 2NEnh]hp^2(x) - 6Enh[M(x,t)+Nh]hp(x) + Enh^2[3M(x,t)+2Nh] = 0

Let's define a new variable, let's say y = M(x,t). This way, we can rewrite the equation as:

2N(Ep-En)hp^3(x) - 3[y(Ep-En) - 2NEnh]hp^2(x) - 6Enh[y + Nh]hp(x) + Enh^2[3y+2Nh] = 0

Now, we have a cubic equation in terms of one variable, x. We can use any method of solving cubic equations to find the solutions. One method is to use the Rational Root Theorem to find possible rational roots, and then use synthetic division to check which of those roots are actual solutions.

However, since y = M(x,t) is still a variable, we cannot find a specific solution for x and t. We can only find a general solution in terms of y.

I hope this helps. If you have any further questions, please do not hesitate to ask. Good luck!