# Orthogonal family of curves

1. Feb 12, 2010

### farleyknight

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

Anyone familiar with orthogonal families of curves? They're not that difficult to understand. If you have a differential equation

$\frac{dy}{dx} = F(x, y)$

you can find it's orthogonal family of curves by solving for

$\frac{dy}{dx} = \frac{-1}{F(x, y)}$

2. Relevant equations

The problem I'm given is to find the orthogonal family for

$y = x - 1 + c e^{-x}$

3. The attempt at a solution

It's fairly easy to see that

$y' = 1 - c e^{-x}$

however, we must eliminate c from this equation. If anyone knows why, that'd be awesome, but that's just the steps that the book mention. So we have

$c = \frac{y - x + 1}{e^{-x}}$ or similarly $-c e^{-x} = x - y - 1$

and then we'd have

$y' = 1 + x - y - 1 = x - y$

so taking the negative reciprocal, to solve for the orthogonal family

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

Now, as far as I know, this equation cannot be solved with elementary functions. I tried a couple of different techniques, well the only ones I have with this course, separable variables, exact equation, linear.. Doesn't seem to fit any of them. When I turn it into a homogenous equation, I don't get a separable equation..

To top it off, to make sure I wasn't crazy, I tried it out in Mathematica and it tells me

DSolve[y'[x] == 1/(y[x] - x), y[x], x]

{{y[x] -> 1 + x + ProductLog[-E^(-1 - x) C[1]]}}

So I don't know WTF.. I'm waiting to look at the back of the book until someone gives me a hint.

2. Feb 12, 2010

### Dustinsfl

If you take the partial of M with respect to y, you should obtain 1, and the partial of N with respect to x it 0; therefore, the partial of M doesn't equal the partial of N. So when you tried the exact method did you then do substitution to make the equation exact since in its current form it is nonexact?

3. Feb 12, 2010

### farleyknight

Well, by letting x be the dependent variable and y be the independent, I managed to come up with this:

$\frac{dy}{dx} = \frac{1}{y - x}$
$\frac{dx}{dy} = y - x$

From this we get the linear equation

$x' + x = y$

which is solved as

$y e^y = \frac{d}{dy} \left( x e^y \right)$
$y e^y - e^y = x e^y + C$
$y - 1 - Ce^{-y} = x$

which, as you might notice, is the same equation with the variables swapped. Still not sure, I plotted the equations side by side and found:

ParametricPlot[
Table[{{t, t - 1 + i E^(-t)}, {t - 1 + i E^(-t), t}}, {i, 0, 2,
0.1}], {t, -1.5, 1.5}]

http://study.farleyknight.com/images/0000/0222/orthogonal_formulas.png?1266025357 [Broken]

So they seem to be roughly orthogonal. I think this might be one of the solutions to this problem, but not sure if it's the one they're looking for.

Last edited by a moderator: May 4, 2017
4. Feb 12, 2010

### Dustinsfl

I obtained your same results as well.

5. Feb 13, 2010

### HallsofIvy

Staff Emeritus
Because a specific value of c designates a specific curve in the family. The equation giving the entire family must not depend on c.