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?

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- Jul 14th 2006, 02:52 PM #1

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- Jul 15th 2006, 10:23 AM #2
I don't think so

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

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

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

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

so

$\displaystyle y(x)=a+(x-1)(1+\frac{1}{a})+... $

- Jul 15th 2006, 01:38 PM #3

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- Jul 15th 2006, 04:19 PM #4

- Jul 15th 2006, 04:25 PM #5

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- Jul 15th 2006, 06:19 PM #6

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I did try to solve this but alas I am having little success.

Rewrite,

$\displaystyle y'-y^{-1}=x^{-1}$

Note, this is a*non-homogenous*.

Thus, begin by finding the general solution of,

$\displaystyle y'-y^{-1}=0$

Thus,

$\displaystyle y'=y^{-1}$

Note, this is a*Bernoulli, equation*

Use substitution,

$\displaystyle u=y^2$

Thus,

$\displaystyle \frac{du}{dx}=\frac{du}{dy}\cdot \frac{dy}{dx}$.

Thus,

$\displaystyle \frac{du}{dx}=2y(y^{-1})=2$

Thus,

$\displaystyle u=2x+C$

Thus,

$\displaystyle y^2=2x+C$

Thus,

$\displaystyle y=\sqrt{2x+C}$---> 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 #7

- Jul 16th 2006, 06:12 AM #8

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- Jul 16th 2006, 06:41 AM #9

- Jul 16th 2006, 08:03 AM #10

- Jul 16th 2006, 08:21 AM #11

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