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how to solve this differential equation

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sri satya ravi
sri satya ravi am 10 Jun. 2016
Kommentiert: Star Strider am 10 Jun. 2016
I want to solve the equation iteratively for rcfv. I have the value of gama = 1.399 and beta = 0.675. How can I solve this equation to get the value of rcfv.

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Star Strider
Star Strider am 10 Jun. 2016
This will get you two real values for ‘r_CFV’:
gama = 1.399;
beta = 0.675;
f = @(rcfv) rcfv.^(1-gama)/gama + ((gama-1)/2)*beta.^4 * rcfv.^(2/gama) - (gama+1)/2;
r_cfv(1) = fzero(f, 1)
r_cfv(2) = fzero(f, 10)
x = linspace(-20, 20);
figure(1)
plot(x, f(x))
grid
Use the Optimization Toolbox fsolve function with real and complex initial estimates to get complex values. Another option is the Symbolic Math Toolbox solve function.
  2 Kommentare
sri satya ravi
sri satya ravi am 10 Jun. 2016
@(rcfv) rcfv.^(1-gama)/gama + ((gama-1)/2)*beta.^4 * rcfv.^(2/gama) - (gama+1)/2; .. Can I know what @(rcfv) at the starting of the equation do. and also can you please explain r_cfv(1) = fzero(f, 1) this equation.
Star Strider
Star Strider am 10 Jun. 2016
I created an anonymous function (see the section on Anonymous Functions in Function Basics) from your equation. The ‘@(rcfv)’ creates a function handle.
Since I expressed your equation as an implicit function, I used the fzero function to find both real roots (there appear to be only two when I plotted your function) at:
r_cfv =
277.0523e-003 8.5921e+000
As I mentioned in my original Answer, there may be complex roots as well that you would have to use the Optimization Toolbox fsolve function or the Symbolic Math Toolbox solve function to find. (When I tried it with the Symbolic Math Toolbox, it only returned the one real solution at 8.5921.)

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adi kul
adi kul am 10 Jun. 2016
Why iteratively when you have only 1 unknown variable?

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