Expert: Paul Klarreich - 12/5/2006

Question
How do you solve the equation du/dx=1-u(1+ln(x))?

Answer
Questioner:  Ken
Subject:   Separable differential equations.

How do you solve the equation du/dx=1-u(1+ln(x))?
.......................................................
Hi, Ken,

Did you mean to write:  du/dx=(1-u)(1+ln(x))?

Well, you didn't write that, so I'll try to solve the equation as you sent it.

Suppose the D.E. said:

du/dx= -u(1+ln(x))

Then I think you would separate the variables this way:
du
--- = -(1 + ln x) dx
u

and integrate each side.  The left side integrates easily to ln u.

For the right side, there are two terms, and the first terms integrates easily to x.

The second term integrates like this:

{
| ln x dx
}
Use Integration by Parts:

u = ln x
du = 1/x dx

dv = dx
v = x

{
| ln x dx =
}
        {
x ln x - | x (1/x) dx =
        }

        {
x ln x - | dx =
        }

x ln x - x

So the right side integrates to:

-(x + x ln x) - x = - x ln x

So we are up to:

ln u = - x ln x + c1

u = exp(- x ln x + c1)

u = c2 exp(- x ln x)

But  exp(x ln x) = [exp(ln x)]^-x = x^-x

u = c2 x^-x

OK, now what about the '1' that we sort of forgot about?  It integrated to x, and it forms another term in the solution:

u = x - c2 x^x

Now I think there might be some more work, but this may be enough for you.