D/dx(dy/dx) + d/dy(dx/dy) = 1

1. Jul 3, 2010

Unit

1. The problem statement, all variables and given/known data
Solve for either x or y:

$$\frac{d^2y}{dx^2} + \frac{d^2x}{dy^2} = 1$$

2. Relevant equations
I don't know any.

3. The attempt at a solution
A little bit of simplification first:

$$\frac{d}{dx}\frac{dy}{dx} + \frac{d}{dy}\frac{dx}{dy} = 1$$

does not seem to help.

Re-writing it as y'' + x'' = 1 does not inspire any further direction either. Actually, it's also ambiguous, because it could mean y''(t) + x''(t) = 1 which is not what I wanted.

I tried letting u = dy/dx and v = dx/dy which also equals 1/u, like this:

$$\frac{du}{dx} + \frac{dv}{dy} = 1$$

Then, reworking v = dx/dy into 1/v = dy/dx, implicit differentiating to get $\frac{-1}{v^2}\frac{dv}{dx} = \frac{d^2y}{dx^2} = \frac{du}{dx}$, removing dx and integrating, only got me back to 1/v = u + C. Hmph. I'm stumped.

2. Jul 3, 2010

Bohrok

[STRIKE]I would try it the same way you did with the other differential equation, rewrite it first as
y'' + 1/y'' = 1

Turn it into a quadratic equation and take it from there.[/STRIKE]

Last edited: Jul 3, 2010
3. Jul 3, 2010

Hurkyl

Staff Emeritus
That's not the same differential equation. While dy/dx dx/dy is one, d²y/dx² d²x/d²y is not.

4. Jul 3, 2010

Bohrok

That notation with higher derivatives gets me sometimes. I see

5. Jul 3, 2010

Unit

Bohrok I thought of that at first too, but Hurkyl is right (although there might be a typo with the second denominator). And I still haven't thought of anything :/

6. Jul 3, 2010

jackmell

Let:
$$u=\frac{dy}{dx}$$,

then

$$1/u=\frac{dx}{dy}$$

so that:

$$u'+\frac{d}{dx}\left(\frac{1}{u}\right)=1$$

isn't that just:

$$u'-\frac{1}{u^2}u'=1$$

Can you finish it? Also, I'm not entirely sure about this.

7. Jul 3, 2010

Unit

jackmell, if 1/u = dx/dy then wouldn't I have to take d/dy of 1/u?

8. Jul 3, 2010

jackmell

Hi. I think what I did is correct but I'm not entirely sure ok. This is how I'd approach it: Either wait til' someone helps us or just for the moment assume what I wrote is ok and work through it to see what happens. Get an answer. Then back-substitute it into the original expression if possible and see if it satisfies the equality. If it does, then there is a chance what I wrote is correct. Then I would proceed to somehow verify that it is indeed valid. Remember, it's very important in mathematics to try something even if it's turns out to be wrong because often the wrong ones lead you to the right answer if you're patient an tolerant of some failures sometimes. :)

Last edited: Jul 3, 2010
9. Jul 3, 2010

Unit

I can rearrange this into

$$u' = \frac{u^2}{u^2-1}$$

Maybe I could use some trigonometry from here. Indeed I can let cosθ = 1/u.
Then, u = secθ, u^2 = secθ, and tanθ = u^2 - 1. However, the u' from above is u'(x) whereas if I take d/dθ of u = secθ then I'll get u'(θ) = secθtanθ. In other words, I've no clue.

10. Jul 3, 2010

Hurkyl

Staff Emeritus
It's a rational function. You've anti-differentiated rational functions before, haven't you?

11. Jul 3, 2010

Unit

So,

du/dx = u^2 / (u^2 - 1)

Separating variables:

(u^2 - 1) du / u^2 = dx

(1 - 1/u^2) du = dx

Integrating:

u + 1/u = x + C

Multiply through by u to get u^1 + 1 = u(x + C). Should I use the quadratic equation now? It's a mess.

12. Jul 3, 2010

jackmell

I get:

$$u^2-u(x+c_1)+1=0$$

and I would then try the quadratic formula. Gotta go.

13. Jul 5, 2010

Phrak

That should be

$$u'+\frac{d}{dy}\left(\frac{1}{u}\right)=1$$

However, this can be changed to

$$u'+\frac{d}{dx}\left(\frac{1}{u}\right)\frac{dx}{dy}=1$$

14. Jul 5, 2010

Unit

Thanks Phrak! That's what I said:

Then you said
This is exactly the step I was just trying today! Thanks a lot, I'll see where I go from here, recalling that dx/dy = 1/u.

15. Jul 5, 2010

Phrak

Yes. I noticed that as I was posting. I though you might want some confirmation anyway.

16. Jul 5, 2010

Unit

I simplified it down to a cubic, u^3 - u^2 (x+C) + 1 = 0. Any suggestions on how to proceed?

17. Jul 5, 2010

Phrak

I don't think that's correct. What do you have for a first order differential equation in u?

18. Jul 5, 2010

Unit

Here is all the work so far:

$$\frac{d^2y}{dx^2} + \frac{d^2x}{dy^2} = 1$$

$$\frac{d}{dx}\frac{dy}{dx} + \frac{d}{dy}\frac{dx}{dy} = 1$$

Let
$$u = \frac{dy}{dx}$$

So the expression becomes
$$\frac{d}{dx} u + \frac{d}{dy} \left(\frac{1}{u}\right) = 1$$

$$u' + \frac{d}{dx}\left( \frac{1}{u}\right) \cdot \frac{dx}{dy} = 1$$

$$u' - \frac{1}{u^2} \cdot \frac{du}{dx} \cdot \frac{1}{u} = 1$$

$$u' - \frac{u'}{u^3} = 1$$

That is my differential equation in u. I can common factor u' to get
$$u' \left( 1 - \frac{1}{u^3} \right) = 1$$

And re-writing u' as du/dx,
$$du \left( 1 - \frac{1}{u^3} \right) = dx$$

Then we can integrate
$$\int 1 - \frac{1}{u^3} du = \int dx$$

$$u + \frac{1}{2u^2} = x + C$$

Multiply through by 2u^2
$$2u^3 + 1 = 2u^2(x + C)$$

$$2u^3 - 2u^2(x + C) + 1 = 0$$

So I admit my expression in post #16 was wrong. Is there anything wrong with my math above?

Last edited: Jul 5, 2010
19. Jul 5, 2010

HallsofIvy

There is a reason for that notation! While $dy/dx$ can be treated "like" a fraction, $d^2y/dx^2$ cannot.

20. Jul 5, 2010

jackmell

Ok. Sorry about that Unit. Been workin' on it but haven't come up with a solution I can successfully back-substitute into the original expression and confirm it satisfies it. Perhaps I can with the result from the cubic.

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