# Homework Help: Differential Equation Where Y Turns Out to Equal X?

1. Jun 17, 2017

### Staff: Mentor

1. The problem statement, all variables and given/known data

Suppose that $$xf(x,y)dx+yg(x,y)dy=0$$

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

2. Relevant equations

3. 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 $dy$?

2. Jun 18, 2017

### haruspex

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

3. Jun 18, 2017

### Staff: Mentor

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

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

4. Jun 18, 2017

### 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?

5. Jun 18, 2017

### Staff: Mentor

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

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

6. Jun 18, 2017

### haruspex

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.