College Math Teaching

March 6, 2010

Implicit Differentiation and Differential Equations

In my “business calculus class”, we were studying implicit differentiation.
We had a problem:
Find dy/dx if x/y - x^2 = 1
I showed them three ways to do the problem, all of which yield different looking answers:

x - x^2y = y Differentiate both sides: 1 - 2xy -(x^2)dy/dx = dy/dx which yields:
dy/dx = (1-2xy)/(1 + x^2)

Method 2: do directly:
(y - (x)dy/dx)/y^2 -2x= 0  This leads to dy/dx = (y - 2xy^2)/x

Of course that looks different; but we can always solve for y and do it directly:
x/y = 1+ x^2 which yields y = x/(x^2+1) which yields the easy solution:
dy/dx = (1 - x^2)/((x^2 + 1)^2

Now one can check that all three solutions are in fact equal on the domain y \neq 0

But here is the question that came to mind:  in the first two methods we had two different two variable equations: (y - 2xy^2)/x^2 = dy/dx = (1-2xy)/(1 + x^2)

So what does this mean for y ?  Is it uniquely determined?

Answer:  of course it is: what we really have is f(x,y)=g(x,y) = dy/dx whose solution IS uniquely determined on an open rectangle so long as f and g are continuous and  \partial f/y and \partial g/y are continuous also. 🙂

But I didn’t realize that until I took my morning swim. 🙂

This is the value of talking to a friend who knows what he/she is doing: I was reminded that f(x,y)=g(x,y) = dy/dx means that f = dy/dx and g = dy/dx indeed have unique solutions that have the same slope at a common point, but with just this there is no reason that the solutions coincide over a whole interval (at least without some other condition).

So we have something to think about and to explore; I don’t like being wrong but I love having stuff to think about!

Now, of course, we have “different” differential equations with the same solution; yes, there is a theory for that.  I’ve got some reading to do!


1 Comment »

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