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

[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.