The DE's that come up in Calculus are Separable. As we just saw this means they can be

written as

and this can be reduced directly to an integration problem

Now how about the DE

or

?

Do you think either of these can be transformed into Separable equations . . .? Try it.

You can be as clever as Ulysses and work like a Trojan but will not be able to pull the

variables apart. We have to turn up the DE voltage to handle such challenges.

Notice that any DE y' = f(x,y) can be written in the form

These more general DE's will require a "back-door" approach. So be a little patient as

we develop the method.

The DE

is Exact

means there is a function u(x,y) with differential

and

.

Such a du is called an "Exact", "Perfect" or "Total" differential.

As we will see in Orthogonal Trajectories (1.8), the expression represents

a one-parameter family of curves in the plane. For example,

is a family of circles of radius and

is a family of parabolas.

Now for a crucial step that involves cross-partials --

Since and ,

and

If we make the mild assumption that

, are continuous

then we get as a freebie that

Now we can carry out two "partial" integrations:

Notice that the integration so-called constants each depend on one of the variables.

Now we do some "criss-crossing" to get our solution .

First, get by solving for dk/dy in

Analogous steps can be carried out for .

Suppose we want to solve the DE

First let's check to see if it is Separable. Well, no, but maybe it is it Exact ...?

It is already in the form

so let's check the "cross-derivatives" for equality --

and .

They are equal, so the equation is Exact! Now to find .

As in the above derivation, we start with

.

Now integrate "partially" with respect to x

so that we have an expression for the solution

However, us is only masquerading as a solution -- the function k(y) is unknown.

But not for long, since we know that

How does that help …? Well, now we can take the partial derivative of the pseudo-

and equate the two expressions for to get

Solving for k'(y) yields

Substitute this known value of k in the pseudo-solution to get

So the general solution is

.

This can be re-written in several ways, for example --

.

Let's check the solution via differentials --

or

.

and this is the DE we started with.