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Gavin's DiffEq Class Clarification: Jan 24

Question: What was that about nullclines?
The book reports, in slightly different words, that a nullcline is a line where solutions to the differential equation have zero slope. The book actually says that nullclines are ``curves of zero inclination''---notice that this is the same thing. So, if we want to find the nullclines for a differential equation
y'(x) = f(x,y),
we have to find where f(x,y) = 0. Let's look at an example: Suppose
y'(x) = x2 - y.
Then nullclines are where x2 - y = 0, so y = x2. (This gives only one nullcline.) Notice that this isn't a solution, because it would have to have to satisfy the differential equation: here y'(x) = 2x, so if we plug into the differential equation we get
2x = x2 - x2 = 0,
which isn't true for all x, so y = x2 isn't a solution to the differential equation.

However, if we had found that a nullcline that was a constant, it would also be a solution curve for an equilibrium solution.

Question: How do I do substitution to integrate?

Let's do an example: Let's suppose that Intab[ ] is the integral from a to b, for the purposes of notation. The integral we'll do is
Int[ (2x - 1) sin(x2 - x) ]dx.
We notice that there is a function composition here (function in another function): the x2 - x appears inside the sine function. So let's let w = x2 - x. This turns the integral into
Int[ (2x - 1) sin(w) ]dx.
Then we need to get rid of the dx in the integral. Differentiating w, we get dw / dx = 2x - 1, so dw = (2x - 1) dx. And as luck would have it, that's exactly what's left in the integral! So it becomes
Int[ sin(w) ]dw = -cos(w) + C = -cos(x2 - x) + C.

Of course, this is a really simple example: possible complications could include

A Calc II clarification gives another take on substitution.

Gavin's DiffEq Clarification 000124
Last Modified: Mon Jan 24 23:32:22 CST 2000
Comments to glarose@umich.edu