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*) = *x*^{2} - *y*.
Then nullclines are where *x*^{2} -
*y* = 0, so *y* = *x*^{2}.
(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*) = 2*x*, so if we plug into the
differential equation we get
2*x* = *x*^{2} -
*x*^{2} = 0,
which isn't true for all *x*, so *y* = *x*^{2} 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 *Int*_{a}^{b}[ ]
is the integral from *a* to *b*, for the purposes of
notation. The integral we'll do is
*Int*[ (2*x* - 1) sin(*x*^{2} - *x*) ]*dx*.
We notice that there is a function composition here (function in
another function): the *x*^{2} -
*x* appears inside the sine function. So let's let
*w* = *x*^{2} -
*x*. This turns the integral into
*Int*[ (2*x* - 1) sin(*w*) ]*dx*.
Then we need to get rid of the *dx* in the integral.
Differentiating *w*, we get
*dw* / *dx* = 2*x* - 1, so
*dw* = (2*x* - 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(*x*^{2} -
*x*) + *C*.

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

- if the integrand has a multiple of the differential
*dw*: then we will get a constant times the answer, or - if the integrand has left-over
*x*'s after we've substituted for the function*w*and*dx*: in this case, we might have to solve the equation for*w*to get*x*=*g*(*w*) and substitute the*g*(*w*) to get rid of the*x*'s---*Note that we can't have any*or*x*'s left over in the equation after finishing the substitution, - if the function to choose for
*w*isn't so obvious. In this case, try (1) any function inside another function, or (2) any function inside a square root, raised to a power, or in the denominator of a fraction, or (3) some other logical function.

A Calc II clarification gives another take on substitution.

Gavin's DiffEq Clarification 000124

Last Modified: Mon Jan 24 23:32:22 CST 2000

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