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Reading: chapter 2 of the course text.

Exercises: from the course text, problems 2.1, 2.2, 2.4, 2.5, 2.7, 2.10, 2.11, 2.12 and 2.13.

Someone asked me:

1. Prob 2.1 asks to say what solution methods work for a given 1st order d.e. Does approximate numerical solution basically work for all d.e.'s?

From the numerical section in the book, it seems to indicate that it will work for any first order you can put in the form

x' = f(x,t)

2. I'm confused about the form of the equation that is necessary to determine if an equation is exact. The following example illustrates this:

(assuming the form f(x,t) * x' + g(x,t) = 0)

If x' + x/t -cos5t = 0, then f=1 and g =x/t - cos5t. From this df/dt=0 does not equal dg/dx=1/t, so the eqn is not exact.

However, if the equation is multiplied by t to obtain t*x' + x - tcos5t=0, the f=t and g=x-tcos5t. Now df/dt=1=dg/dx, and the eqn is exact.

I do not understand why just rearranging the equation would cause this. What form must f and g have in order to perform this test of exactness?

The form can change whether or not it is exact. You don't have to check different arrangements. If it's already in the f(x,t) * x' + g(x,t) = 0 form, you can just check that. The subject of integrating factors is one where you can often multiply a first order equation by something to make it integrable, but isn't something you have to worry about for this homework.

3. Does integration of a separable equation give the only/single possible form of a solution. For example, on problems 2.12 and 2.13, both the diff eqns are separable, and it asks if assuming a given form of the solution works. If I integrated the d.e. to find the solution., and this solution had a different form from the one given, would this mean that the one given was not a soln, since the integration of a separable eqn gives the only/single correct form of the solution.

The solution to a differential equation is generally unique. However, there are different ways to represent the same function. For example, for the top half of a circle you could write y = \sqrt(1-x^2) or you could write it implicitly as x^2 + y^2 = 1.