# Ode

• Jul 14th 2006, 02:52 PM
bobbyk
Ode
Has this old saw ever been answered?
Does the differential equation:
dy/dx=1/x+1/y
have a solution in terms of elementary functions?
• Jul 15th 2006, 10:23 AM
Rebesques
I don't think so :(

This doesn't mean we can't solve it - even approximate the solution numerically. :cool:

As I sucked in Lebesgue spaces, I assume $y$ is smooth, and for the sake of well-posedness, $y(1)=a.$

Then $y'(1)=1+\frac{1}{a}, \ y''(1)=-1-\frac{1}{a^2}-\frac{1}{a^3}$ etc,

and a Taylor's expansion sais $y(x)=a+(x-1)y'(1)+\frac{(x-1)^2}{2}y''(1)+...$

so

$y(x)=a+(x-1)(1+\frac{1}{a})+...$
• Jul 15th 2006, 01:38 PM
bobbyk
Ode
Thanks for responding. I appreciate your interest in this question.

Of course you can solve it numerically! But can the numerical solution be
shown to be an ELEMENTARY function? That's the question.
• Jul 15th 2006, 04:19 PM
Rebesques
...Well I already said i don't think so! :o :o :o
• Jul 15th 2006, 04:25 PM
bobbyk
Ode
Yes you did! And thanks again!
• Jul 15th 2006, 06:19 PM
ThePerfectHacker
I did try to solve this but alas I am having little success.

Rewrite,
$y'-y^{-1}=x^{-1}$
Note, this is a non-homogenous.
Thus, begin by finding the general solution of,
$y'-y^{-1}=0$
Thus,
$y'=y^{-1}$
Note, this is a Bernoulli, equation
Use substitution,
$u=y^2$
Thus,
$\frac{du}{dx}=\frac{du}{dy}\cdot \frac{dy}{dx}$.
Thus,
$\frac{du}{dx}=2y(y^{-1})=2$
Thus,
$u=2x+C$
Thus,
$y^2=2x+C$
Thus,
$y=\sqrt{2x+C}$---> general solution.

For the particiluar solution for this ode I had trouble. I went threw many different attempt non of them work. So maybe this is no elementary function which satisfies this particular condition.
• Jul 16th 2006, 05:22 AM
Rebesques
Quote:

non-homogeneous [...] particiluar solution

I don't think that method applies here, because the equation is not linear. :(
• Jul 16th 2006, 06:12 AM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
I don't think that method applies here, because the equation is not linear. :(

That bothered me too, I chose to ignore it.
• Jul 16th 2006, 06:41 AM
Rebesques
Quote:

That bothered me too, I chose to ignore it.

...U are the man!!! :D
• Jul 16th 2006, 08:03 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
That bothered me too, I chose to ignore it.

We'll make a Physicist out of this guy yet! :D

-Dan
• Jul 16th 2006, 08:21 AM
ThePerfectHacker
I only do that when it comes to diffrencial equations.