Homogenous Diff. Eqn

  • Thread starter aznkid310
  • Start date
  • #1
109
0

Homework Statement



dy/dx = (x + 3y)/(x - y)

A) Solve the Differential Eqn
B) Draw a Direction Field and some integral curves. Are they symmetric w/ respect to the origin?

Homework Equations



I believe i solved the equation correctly, but i dont know how to draw the direction fields and integral curves. I tried plotting y v. y' in order to create the resulting direction field and integral curves, but i dont know what it looks like.

*Also, how do I integrate the left side? I just used an online calculator to get the answer, but i would like to know how to solve this

The Attempt at a Solution



A) After dividing by x and substituting v = y/x:

(1 + 3v)/(1 - v) = xv' + v

v' = (1/x)( (1 + 3v)/(1 - v) - v )

* (1-3v)/(1+3v) - 1/v dv = dx/x

After integrating and solving for c:

C = (2/3)ln(3y/x + 1) - y/x -ln(y/x) - ln(x)
Also, y = -x
 

Answers and Replies

  • #2
benorin
Homework Helper
Insights Author
Gold Member
1,257
71
* Use long division: [tex]\frac{1-3v}{1+3v}=-1+\frac{2}{1+3v}[/tex] then

[tex]\int \left(\frac{1-3v}{1+3v}-\frac{1}{v}\right)\, dv = \int \left(-1+\frac{2}{1+3v}-\frac{1}{v}\right)\, dv = -v+{\textstyle\frac{2}{3}} \ln |1+3v| -\ln |v|+C[/tex]​

A direction field (a.k.a http://mathworld.wolfram.com/SlopeField.html" [Broken], see link for a better definition) is a plot of unit vectors with slope determined by the DE (e.g., on our direction field at (5,1) we plot a unit vector with a slope of [itex]y^{\prime}=\frac{5+3}{5-1}= 2[/itex]).

An http://mathworld.wolfram.com/IntegralCurve.html" [Broken] is a particular solution to a differential equation corresponding to a specific value of the equation's free parameters.
 
Last edited by a moderator:
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hey guys! :smile:

Isn't it (1 - v)dv/(1 + v)² ? :confused:
 
  • #4
benorin
Homework Helper
Insights Author
Gold Member
1,257
71
Thanks tiny-tim.
Oops :blushing:...

[tex]x\frac{dv}{dx}=\frac{1+3v}{1-v}-v[/tex]
[tex] \frac{1-v}{(1+v)^2}dv = \frac{dx}{x}[/tex]
 

Related Threads on Homogenous Diff. Eqn

  • Last Post
Replies
3
Views
1K
Replies
5
Views
412
Replies
4
Views
2K
Replies
3
Views
490
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
416
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
2
Views
866
Top