There's this Differential EQuation:

[1]       dy/dx = 1/(xy) + x

that was discussed on the forum and, tho' the fella who posed the problem really meant dy/dx = 1/x*y + x = dy/dx = y/x + x, [1] is more interesting, eh?

>Huh? I don't find either one very interesting.
Then pay attention! Tho' dy/dx = y/x + x is easy, DEQ [1] ain't.

In [1], we let:

[2]       y = z + x² / 2   ... thereby eliminating that "x" term and getting:

[3a]       dz/dx = 1/{ x(z + x² / 2) }

which, for sanitary reasons, we write as:

[3b]       dx/dz = x(z + x² / 2)

or, better still, as

[3c]       dx/dz - zx = x³ / 2

or, better still, as

[3d]       e^-z2/2dx/dz - z e^-z2/2 x = e^-z2/2 x³/2   ... multiplying everything in sight by e^-z2/2

or, better still, as

[3e]       d/dz {e^-z2/2 x} = e^-z2/2 x³/2

Now let:

[4]       u = e^-z2/2 x     so     x = e^z2/2 u

Then [3e] becomes::

[5]       du/dz = e^-z2/2 x³/2 = e^-z2/2 {e^z2/2 u}³ /2 = e^z2/2 u³/2

>zzzZZZ
Wait! Don't you see? DEQ [5] is a separable equation!

We separate the variables, like so:
2 u^-3 du = e^z2/2 dz
and get:

[6]       u^-2 = - ∫e^z2/2 dz

Then, going back to our original variables ...

>Hey! Where's the ubiquitous +C?
∫e^z2/2 dz is an indefinite integral and includes an arbitrary additive constant.
It's just better to leave our answer unadulterated ... with the "C" implied ... like ... uh, implicitly.

Anyway, in terms of the original x and y variables we get ... from [6], using [4]:

[7]       e^z2 x^-2 = - ∫e^z2/2 dz   where z = y - x² / 2.

>Don't you need a relation between x & y? I mean, you got a relation between x & z and ...

Okay! You can regard [2] and [7] as giving parametric equations for x = x(z) and y = y(z) ... or we can write the relation as:

>I like it much better when I see the C.
I thought you would.
>And you're sure that's the correct solution?
Can anyone be sure of anything these days?
>I assume that the hairy integral has a name, like the Hairy Function?
It's related to the Dawson Function
>How nice. Do I need to know that for the final exam?
Go back to sleep.
>zzzZZZ