"Sam" wrote in message <l346sg$r0c$1@newscl01ah.mathworks.com>...
> (x^0.25) (e^x)=(y^0.25)(e^y).
> In this x value is 0 to 0.3 with an interval 0.01
> I got theLeft side equation now.
> I ve to use each value and solve for y. Can anyone help me out!!!!
> I used x=0:0.01:0.3 and w=(x.^0.25).*exp(x) to get LHS.
         
As Nassar has indicated, there is a matlab function in the Symbolic Toolbox that can solve your problem. It is called 'lambertw'. See its documentation at:
http://www.mathworks.com/help/symbolic/lambertw.html
To see how you can use it, first take the fourth power of both sides of your equation and then multiply both sides by 4, and you get the equation:
4*x*exp(4*x) = 4*y*exp(4*y)
If you then call w = 4*y and z = 4*x*exp(4*x), you arrive at the Lambert W function which expresses w as a function of z:
w*exp(w) = z
This accounts for the computation performed by Nassar.
However, you should realize that in the range of your numbers, there are two real branches of solutions. That is, for a given x there are two possible values of y that satisfy the equation. Nassar called on
lambertw(z)
which gives values in the range 1 <= w <= 0 and therefore 0 <= y <= 1/4. If you call on
lambertw(1,z)
you will get values in the other branch which will range inf < w <= 1 and therefore 1/4 <= y < inf. It is up to you which branch you want your values to be taken from as solutions for y. You may want to use values from the "1" branch for values from 0 to .25 and values from the other branch from .25 to .3, which would constitute a continuous function differing from the obvious solution y = x.
Roger Stafford
