• Support PF! Buy your school textbooks, materials and every day products Here!

Bernoulli ODE question

  • Thread starter cue928
  • Start date
  • #1
130
0
I am working on the following Bernoulli ODE: 3xy^2 y' = 3x^4 + y^3. I come up with n = -2, so v = y^3 and y' = (1/3)v^(-2/3) v'. My integrating factor was x^-1. I end up with y^3 = X^2 + Cx yet the book has the same thing except X^4 instead of X^2. That makes me think I'm going wrong with multiplying it through but I don't see where.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
hi cue928! :smile:

(try using the X2 icon just above the Reply box :wink:)
My integrating factor was x^-1. I end up with y^3 = X^2 + Cx yet the book has the same thing except X^4 instead of X^2.
it should have worked :confused:

show us how you got the x2
 
  • #3
130
0
It was right, I just made a stupid mistake that led me to think it was something bigger than what it really was.
 

Related Threads on Bernoulli ODE question

  • Last Post
Replies
3
Views
866
  • Last Post
Replies
2
Views
867
  • Last Post
Replies
2
Views
924
Replies
3
Views
2K
Replies
14
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
749
  • Last Post
Replies
2
Views
899
  • Last Post
Replies
5
Views
820
Top