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General solution for homogeneous equation

  • Thread starter magnifik
  • Start date
  • #1
360
0
i am having trouble finding the general solution for the given homogeneous equation:

x2yy' = (2y2 - x2)
which i made into
x2dy = (2y2 - x2) dx

i turned it into the following:
(2y2 - x2) dx - x2 dy = 0
then i used substitution of y = xv and got
(2(xv)2 - x2 - x2v) dx - x3 dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v2 - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v2 - 1 - v)
so...
dx/x - dv/(2v2 - 1 - v) = 0
i am stuck here. i know the first part is ln x, but i'm not sure if the second part is [ln (2v2 - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx4/1 - 2Cx3

plz help. thx.
 
Last edited:

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
13
i am having trouble finding the general solution for the given homogeneous equation:

x2dy = (2y2 - x2) dx

i turned it into the following:
(2y2 - x2) dx - x2 dy = 0
then i used substitution of y = xv and got
(2(xv)2 - x2 - x2v) dx - x3 dv = 0
then i turned this into the following ... i think this may be the part i did wrong
(2v2 - 1 - v) dx - x dv = 0
then the integrating factor was
1/x(2v2 - 1 - v)
so...
dx/x - dv/(2v2 - 1 - v) = 0
i am stuck here. i know the first part is ln x, but i'm not sure if the second part is [ln (2v2 - 1 - v)]/(4v - 1)...if this part is right then i am stuck again here. i don't know how to get to the final solution which is

y(x) = x + Cx4/1 - 2Cx3

plz help. thx.
Better if you write your final solution as

[tex]y(x) = \frac{x+Cx^4}{1-2Cx^3}[/tex]
 
  • #3
360
0
that doesn't help me :\
 

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