solve algebraic equation

[amirhossein20n]

hello friends

i need to solve these equations urgently and extract x and y in function of w , separately

i heed your help

 

[nomigoraya]

You can use matlab or mathematica software tool which are used to solve such equations.
Scilab is also free tool in case you do not have matlab license.

 

[pancho_hideboo]

i need to solve these equations urgently and extract x and y in function of w , separately

It has no solution.

 

[asdf44]

Yeah, you can't have entire functions as unknowns.


In general for solving algebraic functions symbolically Microsoft Mathematics is free and quite powerful.

 

[amirhossein20n]

hi pancho_hideboo

why ??

you mean we cant separate X , y ??

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i do with matlab and maple but exactly im not skillful ,
can you help me?

 

[pancho_hideboo]

you mean we cant separate X , y ??

Obviously no solution, unless A or B are function of w.

 

[nomigoraya]

If A and B are constant I am not getting the reason that why there is no solution in terms of omega?
We have two equations and two variables.
Please tell me why there is no solution as I am getting all curious now

 

[pancho_hideboo]

If A and B are constant I am not getting the reason that why there is no solution in terms of omega?

No.
You are misunderstanding.
Surely read a start of this thread.

 

[amirhossein20n]

not surely , in reality A, B are constant

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hi nomigoraya
im sympathetic to your Point of view , but i baffled

 

[nomigoraya]

I will follow this thread to understand what I misunderstood.
Waiting for the reason if there is no solution.

 

[_Eduardo_]

The product \[X(\omega)Y(\omega)\] has a unique value for each \[\omega\]

Then, never be simultaneously \[A+\frac{1}{\omega^2}\] and \[B+\frac{4}{\omega^2}\], unless A or B are functions of \[\omega\] and \[A(\omega)+\frac{1}{\omega^2} = B(\omega)+\frac{4}{\omega^2}\]

However, you would find only the product, not the factors.

 

[amirhossein20n]

hi _Eduardo_

yes you re right ,w has influence on A and B , but important thing must be remarked is that A,B haven't a simple relation with w .but totally yes we could interpret A(w),B(w)

 

[pancho_hideboo]

but important thing must be remarked is that A,B haven't a simple relation with w.
but totally yes we could interpret A(w),B(w)

This contradicts your following append.

not surely , in reality A, B are constant

You also should surely reconsider a start of this thread by yourself.

A-B has to be equal to 3/(w^2).

So X*Y=B+4/(w^2)=A+1/(w^2).

Simply your two equatios have no meaning.
There is only one equation for X*Y.

 

[nomigoraya]

A-B has to be equal to 3/(w^2).

So X*Y=B+4/(w^2)=A+1/(w^2).

Simply your two equatios have no meaning.
There is only one equation for X*Y.

Thanks a lot for such detailed answer now I can understand what was the tricky part of question