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

Homogeneous differential equation

  • Thread starter magnifik
  • Start date
  • #1
360
0
i have the equation x2y' = (y2 - xy)
which i changed to (y2 - xy)dx + x2dy = 0

i then tried to solve using the substitution y = xv
dy = xdv + vdx
so ..
(x2v2) dx + x2(xdv + vdx) = 0
(v2 - v) dx + xdv + vdx = 0
v2dx + xdv = 0
(1/x)dx + (1/v2) dv = 0
ln|x| + C - (1/v) = 0
y(x) = x/ln|x-A|, where A = eC

my answer is COMPLETELY different from what the solution is supposed to be according to wolfram. they got y(x) = 2x/(2Cx2+1)

what did i do wrong?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
You started going wrong on the second line. You didn't change the sign when you moved (y^2-xy)dx over to left side. Why don't you try that again?
 
  • #3
360
0
woops - fail. thanks for pointing that out.
 

Related Threads on Homogeneous differential equation

  • Last Post
Replies
7
Views
4K
Replies
1
Views
844
  • Last Post
Replies
4
Views
727
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
742
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
707
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
0
Views
1K
Top