Dr. Sarah's Comments and Peer Review for James and Jeremy's 10

Dr. Sarah's Comments

fsolve(3*sin(y)+y-2=0,y,0..1); .5170489637 this is one equilibrium solution, and we can see from the dfieldplot that this is an unstable solution since other solutions above and below it tend away from x=.517....

fsolve(3*sin(y)+y-2=0,y,3..4); 3.774518012 is a stable solution

fsolve(3*sin(y)+y-2=0,y,4..5); 4.929543438 is an unstable solution

2nd part - good, but let me clarify the y=x curve a bit. Notice that y=x (as a curve) is NOT a solution of the de. Let's try plugging it in to the d.e. to see this: Well, if y=x as a curve was a solution, then dy/dx=dx/dx =1. So, the lhs of the de is 1. But the rhs of the de is y^2-yx = x^2-x^2 when substituting y=x, so we get a 0 on the rhs of the de. Since 1 is not equal to 0, y=x is not a solution as a curve of the de.

Let's investigate what happens to a solution with an initial condition that lies on y=x. (Say y(4)=4, or y(20)=20). Our initial point (x,x) is on the dfieldplot, and we know that the solution will not stay along the curve y=x, so let's see where it goes. Now we know that initially, the derivative at the point (x,x) will be 0 via the rhs of the de. Hence, the solution curve will travel horizontally a bit to the right (we can see this on the dfieldplot). Well, once this has occurred, then we are below the y=x curve and so our solutions x value is now larger than our solution's y value. It then follows the behavior of the rest of the solutions that are below the y=x curve - it tends towards 0 because the rhs of the de (y^2 -xy) is negative since x>y. Then, the solution curve ends up decreasing.

Peer Review