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

Differential Equ - sub

  • Thread starter freezer
  • Start date
  • #1
76
0

Homework Statement



[itex]yy' + x = \sqrt{x^2 + y^2}[/itex]


Homework Equations



y = vx
y' = v'x + v

The Attempt at a Solution



[itex]y' + \frac{x}{y} = \frac{\sqrt{x^2 + y^2}}{y}[/itex]

[itex](v'x+v) + v =\frac{ \sqrt{x^2 + (vx)^2}}{vx}[/itex]
 
Last edited:

Answers and Replies

  • #2
epenguin
Homework Helper
Gold Member
3,656
725
Stuck? No idea what to do with vx - vx ? Not noticed the whole thing has a factor x?
 
  • #3
76
0
Stuck? No idea what to do with vx - vx ? Not noticed the whole thing has a factor x?
Sorry, i typed the problem wrong the first time.
 
  • #4
epenguin
Homework Helper
Gold Member
3,656
725
Are you still stuck? It seems to me this is fairly easily expressed in terms of y and a variable x/y.
 
  • #5
76
0
Are you still stuck? It seems to me this is fairly easily expressed in terms of y and a variable x/y.
I am sure it is easy one i figure out what i am trying to accomplish with this substition, I am just not certain where it is i am heading for other than d/dx(y') = and something i can integrate on the right side.
 
  • #6
76
0



The Attempt at a Solution



[itex]y' + \frac{x}{y} = \frac{\sqrt{x^2 + y^2}}{y}[/itex]

[itex](v'x+v) + v =\frac{ \sqrt{x^2 + (vx)^2}}{vx}[/itex]


[itex]v' = \frac{ \sqrt{x^2 + (vx)^2}}{vx^2} - \frac{v}{x}[/itex]
 
  • #7
epenguin
Homework Helper
Gold Member
3,656
725
The way I look at it if you write it like this

[itex]y' = -\frac{x}{y} + \frac{\sqrt{x^2 + y^2}}{y}[/itex]

you see the RHS is just a function of x/y. So let this be one of your variables and keep y as the other.
 
  • #8
76
0
The way I look at it if you write it like this

[itex]y' = -\frac{x}{y} + \frac{\sqrt{x^2 + y^2}}{y}[/itex]

you see the RHS is just a function of x/y. So let this be one of your variables and keep y as the other.
Okay, I am a bit dense. I hopefully walking through this will get me going.

v = y/x, x/y = 1/v

[itex]v'x + v = -\frac{1}{v} + \frac{\sqrt{x^2 + (vx)^2}}{vx}[/itex]
 
  • #9
76
0
After talking with another student, they found out this uses v = x^2 + y^2

[itex] v' = 2x + 2yy' = 2(x+yy')[/itex]

[itex]\frac{1}{2} \int \frac{1}{\sqrt{v}} = \int dx[/itex]

[itex]\sqrt{v} = x + c[/itex]

[itex] v = (x+c)^2[/itex]

[itex] x^2 + y^2 = (x+c)^2[/itex]

[itex] y^2 = -x^2 + x^2 + 2xc + C[/itex]

[itex] y = +/- \sqrt{2xc + C} [/itex]
 
Last edited:
  • #10
76
0
The answer in the book is:

[itex] x - \sqrt{x^2 + y^2} = C [/itex]

And that is not what I got :-(
 
  • #11
epenguin
Homework Helper
Gold Member
3,656
725
Yes (9) was well spotted, when you see yy' be looking out for (y2)' .

My thinking was that your d.e. can be written

[tex] y' = - \frac{x}{y} +\sqrt{\left(\frac{x}{y}\right)^2 + 1} [/tex]

and the RHS is purely a function of (x/y) which we could call u, which looked simple. But that is overlooking that then you have to change y' into dy/du - in fact the LHS becomes

Edit something complicated and I think there is no or less than no advantage in this approach which seemed initially attractive after all.
 
Last edited:

Related Threads on Differential Equ - sub

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
721
Replies
4
Views
9K
  • Last Post
Replies
4
Views
723
Replies
8
Views
2K
Replies
21
Views
4K
Replies
5
Views
3K
Replies
1
Views
976
Replies
2
Views
4K
Top