# Homework Help: Can Differential Equations Like This Be Solved?

1. Jul 4, 2012

### Hertz

This isn't a homework problem, but it's a problem I've come across that I would really like to know the answer to. Is it possible?

1. The problem statement, all variables and given/known data

Solve for u(x) and v(x) in the system of equations:

$u(x)v'(x) = x^x(ln(x) + 1)$
$u'(x)v(x) = x^xln(x)$

2. The attempt at a solution

I tried for a long while to solve this, but I have had no differential equations courses outside of the small intro to differential equations section of calc 2.
I could post my work but it would take absolutely forever to type it all up and I really don't think it would be useful considering I sort of just messed around with the differentials not knowing half the time if what I was doing was ok.

Any help is appreciated, and like I said, I don't have much experience with differential equations so please keep your answers as simplified as possible :S

2. Jul 4, 2012

### Curious3141

In principle, yes.

Remember that uv' + vu' = d(uv)/dx, so adding the two equations will give you:

$\frac{d}{dx}(uv) = x^x(1 + \ln x) + x^x\ln x$

Then it's just a matter of integrating both sides. The fly in the ointment in your case is that the RHS does not have an integral in terms of elementary functions. The first term is easily integrable to give $x^x$ but not the second (Mathematica can't do it). So in your example, there's no exact solution.

If you could integrate the RHS, then you have a closed expression for uv as a function of x. Just divide the equation for uv' by the equation for uv to get v'/v, which is a first order separable d.e., which you can solve for v. Similarly, you can solve for u.

Last edited: Jul 4, 2012
3. Jul 4, 2012

### Mute

I think you meant

$$\frac{d}{dx}(uv) = x^x(1 + 2\ln x)$$

4. Jul 4, 2012

### Curious3141

Thanks, I was rushing before. Edited my post.

5. Jul 5, 2012

### jackmell

That's really not enough in my opinion. Suppose Hertz is confronted with this in real-life, a bridge, dam, or something. Does he just stay stuck cus' he can't integrate it nicely?

Hertz, I want you to plug in some numbers, any numbers like say 1 ane 2, and solve it anyway you can and come out with something tangible that you can work with. That is how you "solve" it like you requested.

6. Jul 5, 2012

### Curious3141

For *this* question, yes, he stays stuck. And I have no idea what a bridge or a dam have to do with things.

How would plugging numbers into a pair of simult. d.e.s help him?

But, in the spirit of your suggestion, why not plug in the simple "1" into the RHS of both equations?

You're welcome to show him how to solve uv' = vu' = 1.

Last edited: Jul 5, 2012
7. Jul 5, 2012

### amiras

Ahh u'v = 0 at x=1

8. Jul 5, 2012

### Ray Vickson

People deal with these things every day of the week. That is why powerful numerical methods have been developed. Most problems are unsolvable (in the sense of having nice formulas for their solution) but that does not stop us from dealing with them.

RGV

9. Jul 5, 2012

### klondike

Are you given any initial values or boundary value?

10. Jul 5, 2012

### HallsofIvy

In that case, he would do what any "real-life" engineer would do: integrate numerically. The fact is that "almost all" differential equations cannot be integrated in terms of known functions.

Plug numbers in for what? u and v are functions of x, not parameters. I see no where that you can "plug in some numbers".

11. Jul 5, 2012

### jackmell

I meant initial values for u(x) and v(x), say u(1)=1 and v(1)=2. Now solve it numerically any way you can.

But more importantly, it's gettin' cold and the concrete trucks are at the job site and the guys need to know if they can make a pour, and I tell you what, they don't care about no pretty solution neither. They just want to know if they can pour and you got one hour before the concrete begins setting up in the trucks so get busy solving it and makin' a decision.

. . . ugh, I mean Herts, not you Hall.

12. Jul 5, 2012

### algebrat

Where did you encounter the problem Hertz?

13. Jul 5, 2012

### Hertz

I've had a fascination lately with tetrations and have been attempting to find a way to integrate x^x. This differential equation would have given me a way to do it if it was possible to solve. >.<

14. Jul 5, 2012

### Curious3141

No, I meant u'v = v'u = 1. Identically, meaning that applies for all x.

That was meant to be a very simple example of how to solve this sort of pair exactly. The solution is $u(x) = c_1\sqrt{2x+c_2}, v(x) = \frac{1}{c_1}\sqrt{2x + c_2}$.

I assumed that the OP was asking about exact solutions, because getting a numerical solution generally shouldn't be an issue. Which is why I mentioned (in my very first post): "there's no exact solution".

15. Jul 5, 2012

### Curious3141

I used to play with tetrations and the hyperpower (infinite power tower function) years ago as a mathematical diversion.

There's no way to integrate xx to give a closed form answer in terms of elementary functions. You'll need to write the answer down as the sum of an infinite series, for example.

Many, in fact, most, functions simply cannot be integrated "nicely" - or to be more precise, have integrals that are inexpressible in a finite form in terms of elementary functions. This is provable.

...and when I say "provable", I didn't mean *I* can prove it. I just know it's been done. :tongue2:

Last edited: Jul 5, 2012
16. Jul 6, 2012

### Hertz

Interesting.. Intuitively I would say that the function xx has an obvious region bounded by it and the x-axis. It can be pretty easily seen that the increase in the area of the region continuously changes over the domain. So, I would say that that change in the area should be representable by some sort of manipulation of functions or new functions with new definitions. I'm not sure about the definition of an "elementary function" but if that class of functions does not include xx then surely there are many other functions, maybe yet undefined functions, that are also not included in that class of functions. I would be very interested in seeing a proof that ∫xx dx simply does not exist.

17. Jul 6, 2012

### Curious3141

Not every function expressible as a finite combination of elementary operations/functions has an indefinite integral that can also be so expressed. $x^x$ is one. $e^{-x^2}$ is another. In the case of the latter, the integral is considered so important (to compute areas under the normal curve, for example) that a special function has been defined for it, called $erf(x)$ (more precisely, $\int_0^x e^{-x^2}dx = \frac{1}{2}\sqrt{\pi}erf(x)$). Similarly, you could easily define a special function to equal the integral $\int_0^x x^x dx$ and call it whatever you like.

Proving whether a function has an elementary integral or not is not something I'm familiar with. I do know it has to do with the Risch algorithm.

Last edited: Jul 6, 2012
18. Jul 6, 2012

### Ray Vickson

The basic definitions, theory and proofs of this stuff is obtainable in a more-or-less readable paper by Conrad, at