Differential Equation Where Y Turns Out to Equal X?

[Drakkith]

Homework Statement

[/B]
Suppose that $$xf(x,y)dx+yg(x,y)dy=0$$

Solve: $$f(x,y)dx+g(x,y)dy=0$$

Homework Equations

The Attempt at a Solution

Well, I'm mostly stumbling around in the dark. I tried a few things and got nowhere before heading down this road.

First I solved for $f(x,y)dx$ in the first equation:

$f(x,y)dx=\frac{-yg(x,y)dy}{x}$

I then substituted this into the 2nd:

$\frac{-yg(x,y)dy}{x}+g(x,y)dy=0$

This led to:
$\frac{g(x,y)dy}{g(x,y)dy}=\frac{y}{x}$

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

$y=x$

I'm not sure what to do next or even if I'm going down the right path. I don't really know what this means for the differential equation. Does that turn the entire problem into a function a one variable? If so, does $dx$ turn into $dy$?

 

[haruspex]

Does that turn the entire problem into a function a one variable?

I would read it that x and y are independent by definition. What other solutions are there?

 

[Drakkith]

I would read it that x and y are independent by definition.

So neither X nor Y is a function of the other?

What other solutions are there?

I want to say that whenever $f=-g$ but that seems so obviously trivial.

 

[andrewkirk]

I think the problem may be poorly expressed - ie unnecessarily obscure. The statements 'Suppose that' and the instruction 'Solve' are much too terse and could have multiple meanings.

Perhaps what they were trying to say is as follows:

Given two functions $f,g:\mathbb R^2\to\mathbb R$, and an unknown function $H:\mathbb R^2\to\mathbb R$ such that
$$\frac{\partial }{\partial x}H(x,y)=x\,f(x,y) \textrm{ and } \frac{\partial }{\partial y}H(x,y)=y\,g(x,y)$$
let $\mathscr F_1$ be the family of level sets of $H$. These will be curves in the $x,y$ number plane. With this framework we can re-write the first DE in the form of a total derivative as
$$dH = \frac{\partial }{\partial x}H(x,y)\,dx + \frac{\partial }{\partial y}H(x,y)\, dy = 0$$
which is how we see that the solution is a level set of $H$.

Now consider an unknown function $K:\mathbb R^2\to\mathbb R$ such that
$$\frac{\partial }{\partial x}K(x,y)=\,f(x,y) \textrm{ and } \frac{\partial }{\partial y}K(x,y)=\,g(x,y)$$
and let $\mathscr F_2$ be the family of level sets of $K$. These too will be curves in the $x,y$ number plane, and the second DE can be re-written as:
$$dK = \frac{\partial }{\partial x}K(x,y)\,dx + \frac{\partial }{\partial y}K(x,y)\, dy = 0$$

Find a curve in $\mathscr F_1$ and a curve in $\mathscr F_2$ that intersect, and give the coordinates of the point(s) of intersection.

Take the simplest case first, of $f=g=1$. Then, integrating the second DE we see that $\mathscr F_2$ is the family of straight lines with gradient -1, ie with equation $x+y=C_1$, with parameter $C_1$ indexing the family.

Integrating the first DE we see that $\mathscr F_1$ is the family of circles with equations $x^2+y^2=C_2$, for $C_2\geq 0$, with parameter $C_2$ indexing the family.

We can choose one curve from each family, eg $y=-x$ and $x^2+y^2=1$ and get the two solution points $\pm \left(\frac1{\sqrt 2},\frac1{\sqrt 2}\right)$. But choosing different pairs of curves will give different solutions.

Different choices of $f$ and $g$ will give different sets of possible solutions. Some choices will have no solutions.

Given the multiple possible solutions, maybe that's not what they meant, but I can't see any other natural, theoretically sound interpretation of what they've written.

Is there any more context from the source that may help make sense of their cryptic, oracular signals?

 

[Drakkith]

Is there any more context from the source that may help make sense of their cryptic, oracular signals?

None whatsoever. That is literally the entirety of the problem.

I guess I'll have to ask my professor on Monday. Thanks guys.

 

[haruspex]

I want to say that whenever f=-g but that seems so obviously trivial.

Using that would still give x=y.
I was thinking of f=g=0. Maybe Andrew is onto something, but it seems to me he still ends up with x=y.