Find a function F(x,y) whose level curves are solutions to the differential equation

1. Jan 30, 2006

mr_coffee

Hello everyone!! Its me! I'm stuck, these are suppose to be exact equations, and yet its not in exact equation forum. I thought being an exact equation you have to have the following form:

But my question is the following:
Find a function F(x,y) whose level curves are solutions to the differential equation

I rearranged it so its now:
y - x dx/dy = 0
but its not dy/dx, its dx/dy! so what happens? T hanks.

2. Jan 30, 2006

Integral

Staff Emeritus
It means that you would be looking for some function
x(y)

You could just as well express it as

$$x - y \frac {dy} {dx} = 0$$

if it makes you more comfortable.

3. Jan 30, 2006

wms121

..test some simple solutions first

For instance:

Let M(x,y) = x^n A(y) and N = x^m B(y), where y = (sum over all i's) of:

r0 + r1 x + r2 x^2 + r3 x^3 ....r(i)x^i for i less than I(sub)n.

See if M=constant works..N=constant works...etc..

Then try some transcendentals..exp(x), sin(x) etc..

Non-linear equations want to you to 'play with them'..before they tell
you what is going on.

WW

<---getting LaTex soon

4. Jan 30, 2006

mr_coffee

Thanks, so what ur saying is i just gota start guessing? how do uknow what to even guess by?

5. Jan 30, 2006

wms121

simple solutions (part II)

y = +/- sqr(x^2) + C0 when multiplied by a second solution ys = W(x)s

where W = {(sum over all j) v0 + v1 x + v2x^3 ...v(j)x^j} times s(y),

may give more solutions for your N= (whatever) ; M(whatever) non-polynomial factors. Some simple polynomials however lead directly into very involved transcendentals..for instance the D.E.: (second DE)

y" = x^2 + x is a form of the elliptic logarithm.

You might try the Steven Wolfram site on an "unsolvable".

WW

6. Jan 31, 2006

HallsofIvy

Staff Emeritus
Why would you rearrange it like that? You don't have to have any particular form in order for an equation to be exact. "exact" simply means that it can be put in the form M(x,y)dx+ N(x,y)dy = 0 and that there exist a function F(x,y) such that dF(x,y)= M(x,y)dx+ N(x,y)dy so that the equation is of the form dF(x,y)= 0 and has solution F(x,y)= C. One way to check for that is to check the equality of the mixed second derivatives of F: Fxy= My= Nx= Fyx.

ydy- xdx= 0 certainly is exact: (y)x= 0= (-x)y. You could then say: Fy= y so F(x,y)= (1/2)y2+ f(x). From that Fx= f'(x)= -x so f(x)= -(1/2)x2. The solution is F(x,y)= (1/2)(y2- x2)= C

Even more simply, the equation is obviously separable: ydy= xdx. Integrating both parts, (1/2)y2= (1/2)x2+ C which gives exactly the same answer.

7. Jan 31, 2006

wms121

..ok..correction

dy^2/dx^2 = y + y^2 (elliptic logarithm)

..and,...hmm they use these as basis spaces for Complex Proofs..

Have you had numerical series solving? There are existance proofs
using summation analysis, product functionals and Fourier.

For instance the 2D Laplace form for F(x,y):

L(f,s) = int f(x,y) exp(-sy) dy int exp(-vx)dx | from 0 to ymax:xmax

I am getting LaTex before I come back here.

WW

*(..cancelling my warp drive journal..till then as well..email me for info)

8. Jan 31, 2006

mr_coffee

Thanks everyone, and thank you IVEY, i don't know why I thought it had to be in that form. That was alot easier then I expected! and yes the answer was: