Differential equation first, first degree help

  • Thread starter elphin
  • Start date
  • #1
18
0
Differential equation first, first degree help!!

Homework Statement



solve: x.dy + y.dx + (x.dy - y.dx)/(x^2 + y^2) = 0

Homework Equations



if M.dx + N.dy = 0 has to be exact then

equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

The Attempt at a Solution



the idea is to find if this equation is exact because once you do that the integration is easy..

but

now i first i simplify the equation by multiplying throughout by (x^2 + y^2)

simplified form: (x^3 + x.y^2 - y).dx + (y^3 + y.x^2 + x).dy = 0
try to check if it satisfies equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

here M = (x^3 + x.y^2 - y) & N = (y^3 + y.x^2 + x)

we get 2.x.y - 1 is not equal to 2.x.y + 1

now i try another method i.e. group the terms without multiplying throughout by (x^2 + y^2)

simplified equation: [x - (y/(x^2 + y^2))].dx + [y + (x/(x^2 + y^2))].dy = 0
now when we apply equation 1 criteria we get:

partial derivative of M w.r.t y (keeping x constant) = (y^2 - x^2)/(x^2 + y^2)^2 = partial derivative of N w.r.t x (keeping y constant) = (y^2 - x^2)/(x^2 + y^2)^2

my question is - why is it that i am getting the two methods to be different???
 

Answers and Replies

  • #2
18
0


eh... anyone???? wold really help.. this has been bugging me...
 
  • #3
525
16


Have you seen integrating factors yet? That whole method is based on the idea of multiplying the equation by a function to put it in a 'nice' form. When you have an equation M(x,y)dx + N(x,y)dy=0, multiplying by anything is going to change the derivatives. It's very likely that it will change the partial derivatives of M and N in different ways, which may or may not help you make the equation into an exact equation. For example, if I multiply the equation by x, that's going to change the x partial derivative of N, but it's not going to change the y partial of M.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,368
1,035


Homework Statement



solve: x.dy + y.dx + (x.dy - y.dx)/(x^2 + y^2) = 0

Homework Equations



if M.dx + N.dy = 0 has to be exact then

equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

The Attempt at a Solution



the idea is to find if this equation is exact because once you do that the integration is easy ...
Notice that, d(xy) = x.dy + y.dx .

Also, d(y/x) = (xdy - ydx)/x2

So, you can rewrite your equation as: (x2 + y2)d(xy) + x2(d(y/x)) = 0

This suggests that you let u = xy and v = x/y.

Can you take it from here?
 

Related Threads on Differential equation first, first degree help

Replies
5
Views
1K
Replies
1
Views
936
Replies
2
Views
2K
  • Last Post
Replies
2
Views
731
Replies
5
Views
589
  • Last Post
Replies
3
Views
679
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
822
  • Last Post
Replies
3
Views
1K
Top