Help using fzero function in matlab

[Pepsi24chevy]

Hey guys i have nonlinear equation: dy/dt = 3siny + y - 2. I got the direction field with this

>> [T,Y] = meshgrid(-5:00.2:5, -5:0.2:5);
>> S = (3*sin(Y))+Y-2;
>> L = sqrt(1 + S.^2);
>> quiver(T, Y, 1./L, S./L, 0.5), axis tight
>> xlabel 't', ylabel 'y'
>> title 'Direction Field for dy/dt = (3*sin(Y))+Y-2'

NOw i need to use fzero function to find approximate values for all the equilibria of teh system. I just dont' know how to use fzero with a dif equation like this. ANy help is appreciated. thanks Also, i know the values are around 0.5 and 4.

 

[saltydog]

Hey guys i have nonlinear equation: dy/dt = 3siny + y - 2. I got the direction field with this

>> [T,Y] = meshgrid(-5:00.2:5, -5:0.2:5);
>> S = (3*sin(Y))+Y-2;
>> L = sqrt(1 + S.^2);
>> quiver(T, Y, 1./L, S./L, 0.5), axis tight
>> xlabel 't', ylabel 'y'
>> title 'Direction Field for dy/dt = (3*sin(Y))+Y-2'

NOw i need to use fzero function to find approximate values for all the equilibria of teh system. I just dont' know how to use fzero with a dif equation like this. ANy help is appreciated. thanks Also, i know the values are around 0.5 and 4.

The equilibrium points for an autonomous DE are the values of y such that the RHS=0. If I plot the RHS as a function of y, I get 3 zeros. I use Mathematica so can't help with matlab. Doesn't it have a help function? You know, place the cursor on the function then press the F1 key. Surely you can just treat it (RHS) as an algebraic equation in y and solve for the roots right?

 

[tacman]

Hi there,

To use the fzero function in MATLAB, you will need to define a function handle for your nonlinear equation. In this case, your equation is dy/dt = 3siny + y - 2, so you can define a function handle like this:

f = @(y) 3*sin(y) + y - 2;

Next, you will need to choose an initial guess for the root of your equation. This is where the equilibrium values will be located. Since you already have a direction field, you can use that to estimate the equilibrium values. For example, from your direction field, it looks like there is an equilibrium value at around y = 0.5 and another one at y = 4. So, you can use these as initial guesses for your fzero function.

To find the first equilibrium value, you can use the following code:

x1 = fzero(f, 0.5);

This will give you the value of x (or y in this case) where the function f is equal to 0. In other words, it will give you the first equilibrium value.

To find the second equilibrium value, you can use the same code but change the initial guess to 4:

x2 = fzero(f, 4);

You can repeat this process for any other equilibrium values that you see in your direction field. Just make sure to change the initial guess to the approximate location of the equilibrium value.

I hope this helps. Let me know if you have any other questions. Good luck!