What is the independent variable in the equation d^2y/dx^2 + d^2x/dy^2 = 1?

[Unit]

Homework Statement

Solve for either x or y:

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

Homework Equations

I don't know any.

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.

 

[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]

 

[Hurkyl]

y'' + 1/y'' = 1

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

 

[Bohrok]

That notation with higher derivatives gets me sometimes. I see

 

[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 :/

 

[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.

 

[Unit]

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

 

[jackmell]

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

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. :)

 

[Unit]

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

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.

 

[Hurkyl]

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

 

[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.

 

[jackmell]

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.

I get:

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

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

 

[Phrak]

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

then

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

so that:

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

That should be

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

However, this can be changed to

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

 

[Unit]

That should be

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

Thanks Phrak! That's what I said:

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

Then you said

However, this can be changed to

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

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.

 

[Phrak]

Thanks Phrak! That's what I said:

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

 

[Unit]

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

 

[Phrak]

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

 

[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?

 

[HallsofIvy]

That notation with higher derivatives gets me sometimes. I see

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

 

[jackmell]

That should be

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

However, this can be changed to

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

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.

 

[Dickfore]

you need to express the second derivative:

$$\frac{d^{2} x}{d y^{2}}$$

In terms of derivatives of y(x) with respect to x. For this you will need the chain rule. Let me show you the first derivative:

$$\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \frac{1}{y'}$$

 

[Dickfore]

Since the above discussion went further than where I stopped, I think it is safe to write the second derivative as well:

$$\frac{d^{2} x}{dy^{2}} = \frac{d}{d y}\left(\frac{d x}{d y}\right) = \frac{d x}{d y} \, \frac{d}{d x} \left(\frac{d x}{d y}\right)$$

$$\frac{d^{2} x}{dy^{2}} = \frac{1}{y'} \, \frac{d}{d x} \left(\frac{1}{y'}\right) = \frac{1}{y'} \, \left( -\frac{1}{(y')^{2}}\right) \, y'' = -\frac{y''}{(y')^{3}}$$

Then, your equation reads:

$$y'' - \frac{y''}{(y')^{3}} = 1$$

You can reduce the order by one by introducing the substitution:

$$z = y'$$

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

This equation can be integrated now.

 

[Unit]

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

This equation can be integrated now.

Thanks Dickfore! I got this far too, on the same chain rule principle, where my u = your z = dy/dx:

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

That is my differential equation in u.

Then, as you suggest, I integrated:

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

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

The problem is, I don't know where to go from here. Tartaglia's method for finding a real root of a cubic equation does not look inviting...

 

[Dickfore]

The integrated equation is easily solved in terms of x as a function of u (z). Thus, your solution is given in parametric form x = x(u). You still need to find y = y(u). For this, use:

$$\frac{d y}{d u} = \frac{d y}{d x} \, \frac{d x}{d u} = u \, \frac{d x}{d u}, \ u = y' = \frac{dy}{dx}$$

and the already known equation for x = x(u) to express dx/du. Then, integrate again.

 

[Unit]

Thanks, Dickfore. I didn't realize I could express my solution in parametric form x = x(u), y = y(u), and u = dy/dx.

Following your hint, I did dy/du = u dx/du = u (1 - 1/u^3) = u - 1/u^2. Integrating, y(u) = 1/2 u^2 + 1/u + D. I picked D instead of C to distinguish it from the constant in x(u).

I don't really see where to go from here, though. Or is this as far as the solution should go?

 

[Dickfore]

This is the solution. You have 2 arbitrary constants C and D, since your original differential equation was of second order.

 

[Unit]

Thanks a lot! I really appreciate your help and I'm learning a lot. Our solution so far is this:

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

$$y(u) = \frac{u^2}{2} + \frac{1}{u} + D$$

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

To check with the original expression: I can find du/dx easily. However, finding d/dy of 1/u is not apparent to me. Do you have any suggestions?

EDIT: Wait, can I take d/dy of y(u) and by implicit differentiation rearrange for d/dy of 1/u?
EDIT 2: YES! IT WORKED! I am happy! Thanks a lot Dickfore and everyone else :)

The only question I have now is, is there any possible way to solve for u?

 

[Bohrok]

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

I know; that's why I never really liked that differentials can usually be treated like fractions, because one day, when one isn't really thinking...

I will say I've learned some great tips in this thread.

 

[Dickfore]

Thanks a lot! I really appreciate your help and I'm learning a lot. Our solution so far is this:

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

$$y(u) = \frac{u^2}{2} + \frac{1}{u} + D$$

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

To check with the original expression: I can find du/dx easily. However, finding d/dy of 1/u is not apparent to me. Do you have any suggestions?

EDIT: Wait, can I take d/dy of y(u) and by implicit differentiation rearrange for d/dy of 1/u?
EDIT 2: YES! IT WORKED! I am happy! Thanks a lot Dickfore and everyone else :)

The only question I have now is, is there any possible way to solve for u?

Yes, it is a 3rd order equation, and those are always solvable in closed form.

 

[Unit]

Yes, it is a 3rd order equation, and those are always solvable in closed form.

How is it a 3rd order equation if the original problem involved only 2nd derivatives? Do you mean it was a polynomial equation of the 3rd degree? That would make more sense.

 

[Dickfore]

How is it a 3rd order equation if the original problem involved only 2nd derivatives? Do you mean it was a polynomial equation of the 3rd degree? That would make more sense.

Yes. I am sorry for the bad terminology.

 

[Phrak]

...Then we can integrate

$$\int 1 - \frac{1}{u^3} du = \int dx$$

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

I don't see anything wrong with this approach. The problem that I am having is that Mathematica delivers a completely different result, that's not at all pretty by the way, involving several lines.

 

[hellofolks]

The integral curves in parametric form seem to be correct. What I don't understand is that the problem asked for a solution in terms of either x or y and this is not the case for the one given. Are you sure you were asked to solve for either x or y? Didn't they just ask for a family of integral curves?

 

[Unit]

The integral curves in parametric form seem to be correct. What I don't understand is that the problem asked for a solution in terms of either x or y and this is not the case for the one given. Are you sure you were asked to solve for either x or y? Didn't they just ask for a family of integral curves?

Actually, the problem was inspired by the textbook I'm currently perusing (Murray R. Spiegel's Applied Differential Equations, 3rd. ed). Question C1 on page 14 reads,

In the equation $dy/dx + dx/dy = 1$, which variable is independent? Which variable is independent in the equation

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

I know that in both of these equations the distinction between dependent and independent variables is blurred. It was out of curiosity that I wanted to find solutions. As you can see in the thread for the first equation, we succeeded to find a y = F(x). I naively assumed it was possible for the second-order equation, too. So that's why my instruction in the first post was "Solve for either x or y"; I made it up.