Solving a Differential Equation: Finding F(x,y)

[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:

http://tutorial.math.lamar.edu/AllBrowsers/3401/Exact_files/eq0008M.gif

But my question is the following:
Find a function F(x,y) whose level curves are solutions to the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/a6/62aa7a4c8735852fb4ffa4df53d26d1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png
I rearranged it so its now:
y - x dx/dy = 0
but its not dy/dx, its dx/dy! so what happens? T hanks.

 

[Integral]

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.

 

[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

 

[mr_coffee]

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

 

[wms121]

simple solutions (part II)

...your general form:

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

 

[HallsofIvy]

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:

http://tutorial.math.lamar.edu/AllBrowsers/3401/Exact_files/eq0008M.gif

But my question is the following:
Find a function F(x,y) whose level curves are solutions to the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/a6/62aa7a4c8735852fb4ffa4df53d26d1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8d/8616ca1dd6ce230911485b98a57fa31.png
I rearranged it so its now:
y - x dx/dy = 0
but its not dy/dx, its dx/dy! so what happens? T hanks.

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: F_xy= M_y= N_x= F_yx.

ydy- xdx= 0 certainly is exact: (y)_x= 0= (-x)_y. You could then say: F_y= y so F(x,y)= (1/2)y²+ f(x). From that F_x= 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.

 

[wms121]

..ok..correction

should read:

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 existence 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

...enough already*.

I am getting LaTex before I come back here.

WW

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

 

[mr_coffee]

Thanks everyone, and thank you IVEY, i don't know why I thought it had to be in that form. That was a lot easier then I expected! and yes the answer was:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/f5/7102c63fed37197e439d0270e34b001.png