Need help setting up diff equation in matlab

[Pepsi24chevy]

The problem reads as followed: Plot the direction field fo rthe equation dy/dt = y^2-ty

again using a rectangle large enough to show the possible limiting behavors. Identify the unique constant solution. Why is this solution evident from the differential equation? If a solution curve is ever below the constant solution, what must its limiting behavor be as t increases? For solutions lying above the constant solution, describe two possible limiting behavors as t increases. there is a solution curv e that lies along the boundary of the two limiting behavors. What does it do as t increases.

ok, now i am having problems plotting the field. I get all the direction vectors pointing down no matter what domain i choose. Here is how i have been typing it in.
>> [T,Y] = meshgrid(-5:0.2:5, -5:0.2:5);
>> S = Y^2 - T*Y;
>> L = sqrt(1 + S.^2);
>> quiver(T, Y, 1./L, S./L, 0.5), axis tight

 

[Pepsi24chevy]

anyone? any help is appreciated

 

[tacman]

To set up the differential equation in MATLAB, you can use the "ode45" function. This function solves a system of differential equations of the form dy/dt = f(t,y) using the fourth and fifth order Runge-Kutta method. In your case, the function f(t,y) would be y^2-ty.

To set up the equation, you can define a function handle for f(t,y) using the "@" symbol and then pass it as an argument to the "ode45" function. For example:

f = @(t,y) y^2 - t*y;
[T,Y] = ode45(f, [tspan], [y0]);

Here, "tspan" is the time interval over which you want to solve the equation and "y0" is the initial value of y at t=0.

To plot the direction field, you can use the "quiver" function. This function plots arrows representing the direction and magnitude of the vector field at each point in the meshgrid. To create the meshgrid, you can use the "meshgrid" function, as you have done in your code. However, in order to use the "quiver" function, you need to pass it the values of the vector field at each point, which can be calculated using the function handle "f". So, your code would look something like this:

[T,Y] = meshgrid(-5:0.2:5, -5:0.2:5);
S = Y.^2 - T.*Y;
L = sqrt(1 + S.^2);
quiver(T, Y, 1./L, S./L, 0.5), axis tight

Note that you need to use the element-wise operators "." and ".*" to perform calculations on the entire matrix.

To identify the unique constant solution, you need to find the values of y for which dy/dt = 0. In this case, it would be when y = 0 or y = t. These values are evident from the differential equation because when y = 0, dy/dt = 0 and when y = t, dy/dt = 0. This means that y = 0 and y = t are constant solutions.

If a solution curve is ever below the constant solution (y = 0 or y = t), its limiting behavior as t increases would be to approach the constant solution. This is because as t