1. Equation solution Matlab

Can any any body help me solve this equation in Matlab
-a2(b4-b2) +a+n(b-b3)+4n{(b2a/n)+b}0.5 (b2.5-1) =1
condition
For n=0
I have to find the value of “a” when b varying from [0,1] with the step of.025
Then same thing for n=1,2,3
PLEASE SEND ME THE SOLUTION ..URGENTLY HELP REQUIRED.. THANKS

2. Originally Posted by anshu19
Can any any body help me solve this equation in Matlab
-a2(b4-b2) +a+n(b-b3)+4n{(b2a/n)+b}0.5 (b2.5-1) =1
condition
For n=0
I have to find the value of “a” when b varying from [0,1] with the step of.025
Then same thing for n=1,2,3
PLEASE SEND ME THE SOLUTION ..URGENTLY HELP REQUIRED.. THANKS
That is incomprehensible, please reword more clearly.

CB

3. Its a implicit function and difficult to separate the the variable

4. Originally Posted by anshu19
Its a implicit function and difficult to separate the the variable
That does not help.

You will not get much help from anyone if your equation cannot be understood. Learn latex. See here: http://www.mathhelpforum.com/math-he...-tutorial.html

Then re-post the equation.

5. Equation is a a implicit function

(a^2)*(b^4-b^2) +a+n*(b-b^3)+4*n*{(b^2*a/n)+b}^0.5 *((b^2.5)-1) =1
condition
For n=0
I have to find the value of “a” when b varying from [0,1] with the step of.025
Then same thing for n=1,2,3
PLEASE SEND ME THE SOLUTION ..URGENTLY HELP REQUIRED.. THANKS

6. Its a simple equation in which there are three variable.
1st case
n=0
b=0
a=?
then again i have to find out
at n=0
b=0.025
a=?
then
n=0
b=.05
a=?
so on

like this way i have to find the value of 'a' at different value of b and n