Need help on a nonlinear first order DE question

  • Thread starter huaxue09
  • Start date
  • #1
huaxue09
1
0
Hi Everyone

I tried all ways I can to solve (y^2+x^2+x)y'-y=0, but still cann't find a way to solve it.

Could anybody help on it?

Thank you very much!
 

Answers and Replies

  • #2
jackmell
1,804
53
Try and get it exact. You can write it as:

[tex](y^2+x^2)dy+xdy-ydx=0[/tex]

Ok, divide by x^2 to get:

[tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

bingo-bango right?
 
  • #3
chief10
78
0
you wouldn't go one further?
 
  • #4
JJacquelin
801
34
Hi !
Another method consists in converting to polar coordinates, which leads to a linear ODE very easy to solve.
The result can be expressed on a parametric form : x=(c-t)/tan(t) , y= c-t
or as x fonction on y :
x = y / tan(c-y)
 
  • #5
jackmell
1,804
53
you wouldn't go one further?

Well, one is a differential of just y and the other is a differential of the quantity y/x. So then if we have:

[tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

we could re-write that as:

[tex]dy=\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

Now, that right side is equivalent to the expression:

[tex]\frac{du}{f(u)}[/tex]

right? So we just integrate it:

[tex]\int dy=\int\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

and get:

[tex]y=\arctan(y/x)+c[/tex]
 

Suggested for: Need help on a nonlinear first order DE question

Replies
8
Views
152
Replies
6
Views
1K
Replies
5
Views
298
Replies
4
Views
583
Replies
12
Views
679
  • Last Post
Replies
4
Views
199
  • Last Post
Replies
0
Views
344
Replies
2
Views
585
  • Last Post
Replies
4
Views
735
  • Last Post
Replies
3
Views
405
Top