As announced in class today, our first midterm is this Wednesday from 7:30-8:20 in UC McConomy. There will be no class on Wednesday. I will hold a review session tonight from 6:30-8:00 in PH 100.
I've added "Systems of Differential Equations (Section 7.1)" to the list of topics covered. That is what I've been telling people all along, but I inadvertently left it off the list. I also added 7.1.23 to the practice problems from the text.
You asked "Why do we need a rectangle for the second theorem?" The short answer is: "because nonlinear equations are not as nice as linear equations." But let me turn the question around: Why doesn't the first theorem require a rectangle?
If we start with the equation y' + p(t)*y = g(t), we rearrange things to get y' = g(t)-p(t)*y. We can apply the second theorem to this equation with f(t,y) = g(t)-p(t)*y.
If p and g are continuous on the interval (a,b), then f is continuous on the rectangle (a,b)x(-infty,+infty).
The partial derivative of f with respect to y is df/dy=-p(t). Since p is continuous on the interval (a,b), when we think of df/dy as a function of t and y, it is continuous on the rectangle (a,b)x(-infty,+infty).
So for linear equations, we can always choose a rectangle that is infinitely tall. There is one other issue, though. The theorem for linear equations says the domain of the solution is all of (a,b). An implication of this is that the solution cannot have a vertical asymptote before t=b. Why is this?
Well, a complete explanation is beyond the scope of this course, but it is basically because the growth of the slope y' only depends linearly on y, it can't grow quickly enough to get to +infty unless p(t) or g(t) has a discontinuity.
Here's a related problem: Consider the differential equation y'=y^p, where p is a constant greater than zero. This is a separable equation, and if you solve it, you will find that if p<=1, then the the solution is defined for all values of t, but if p>1, then the solutions have a vertical asymptote.
