# 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!