How to Solve Non Linear equation with two variables?
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f = c*(p - g)*((t(2) - t(1)) - a*(t(1) - t(2))^(b + 1)/(b + 1)) + g*(t(2) + a*(t(2)^(b + 1)/(b + 1))) + (c*(p - g)/(a*b))*((-b + 3)*(t(1) - t(2))^(-b + 3)/2 - 3*(t(1) - t(2))^2/((b + 1)*(b + 2))) + (g/a*b)*((-b + 3)*t(2)^(-b + 2) + (2*a*t(2)/((b + 1)*(b + 2))));
Here "t" is the variable "a" & "b" varies 0<a<1 and b>0, rest of them are constants
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Walter Roberson
il 23 Nov 2015
I took the liberty of reformatting your "t (1)" and "t (2)" to "t(1)" and "t(2)"
Risposte (1)
Walter Roberson
il 23 Nov 2015
You cannot uniquely solve a single equation in two variables.
You can symbolically solve in terms of t(1), getting out a RootOf() form:
RootOf(c*(b-3)*(b+2)*(b+1)*(g-p)*(Z-t2)^(-b+3) + 2*a^2*b*c*(g-p)*(b+2)*(Z-t2)^(b+1) + (-2*b^5*g+14*b^3*g+12*b^2*g)*t2^(-b+2) + 2*a^2*b*g*(b+2)*t2^(b+1) + 2*a*((g-p)*(Z-t2)*c+g*t2-f)*b^3 + 6*a*((g-p)*(Z-t2)*c+(5/3)*g*t2-f)*b^2 + 4*a*((g-p)*(Z-t2)*c+g*t2-f)*b + (6*(g-p))*(Z-t2)^2*c, Z)
Rootof(f(Z),Z) is the set of values, Z, such that f(Z) = 0 -- the roots of f(Z).
You can see by examination that Z appears raised to the power of b. You indicated that b > 0. If b is the integer 1, 2, or 3, then there may be analytic solutions readily identifiable. If b is an integer > 3 then by multiplying through by Z^(b-3) you might happen to end up with something that has an analytic solution. If b is non-integral then the set of solutions to the equation is going to depend upon how you define a negative value to a fractional power
You are almost certainly going to have multiple roots to the equation, some of which are negative, and you are probably going to get into complex solutions.
If b is integral then you will be able to enumerate a complete set of solutions; if b is not integral then the set of solutions will depend how you define values to non-integral powers; if b is at least to be fundamentally expressed in rational form then you should be able to enumerate all of the solutions, but if you define b in terms of a floating point number or in terms of an irrational number then you have difficulty enumerating solutions and will probably only be able to find a single numeric solution (no chance of analytic solutions in that case.)
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