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

DE Help - PLEASE!

  • Thread starter aatkins09
  • Start date
  • #1
7
0
dy/dx=(-x+sqrt(x^2+y^2))/y

USE SUBSTITUTION u=x^2+y^2

I tried to work it out but got a really ugly answer, please help!
 

Answers and Replies

  • #2
12
0
If you use the correct substitution then several things should cancel. It would be helpful to also post your attempt at the solution in order to see where you are having problems.
 
  • #3
7
0
If you use the correct substitution then several things should cancel. It would be helpful to also post your attempt at the solution in order to see where you are having problems.
I got
int(y dy)=int(-x+sqrt(u))dx
which turned into
y^2/2=(1/-2+2y)*int(sqrt(u) du)
which equals
-1/2+2y*2/3u^3/2
which then gets down to
y^2/2=(-u^(3/2))/3-3y

and then it gets uglier when I try to get y by itself
I know I am messing up somewhere but cannot pin point it
 
  • #4
uart
Science Advisor
2,776
9
Instead of starting out by trying to separate and integrate, try finding an expression for du/dx and using it to replace dy/dx
 
  • #5
7
0
Instead of starting out by trying to separate and integrate, try finding an expression for du/dx and using it to replace dy/dx
du/dx would be 2x+2y but I dont understand where to go from there. how can I replace dy/dx with it?
 
  • #6
12
0
du/dx would be 2x+2y but I dont understand where to go from there. how can I replace dy/dx with it?
Remember that you have to use implicit differentiation.
[tex]du=2xdx+2ydy[/tex]

Then you can solve for [tex]dy/dx[/tex]
You should find that when you find this and substitute, several things should cancel.
 
  • #7
7
0
Remember that you have to use implicit differentiation.
[tex]du=2xdx+2ydy[/tex]

Then you can solve for [tex]dy/dx[/tex]
You should find that when you find this and substitute, several things should cancel.
I did it (in my head, not on paper yet) and I'm pretty sure I've got it! thank you so much, I appreciate all of your help!
 
  • #8
7
0
Remember that you have to use implicit differentiation.
[tex]du=2xdx+2ydy[/tex]

Then you can solve for [tex]dy/dx[/tex]
You should find that when you find this and substitute, several things should cancel.
nvm, I still can't get it. it's okay though, thank you, I appreciate all of your help!
 
  • #9
uart
Science Advisor
2,776
9
nvm, I still can't get it. it's okay though, thank you, I appreciate all of your help!
[tex] \frac{du}{dx} = 2x+2y \, \frac{dy}{dx}[/tex]

Now just substitute in your original expression for dy/dx and it literally just falls into place.
 
  • #10
7
0
[tex] \frac{du}{dx} = 2x+2y \, \frac{dy}{dx}[/tex]

Now just substitute in your original expression for dy/dx and it literally just falls into place.
Thanks so much :)
 

Related Threads on DE Help - PLEASE!

  • Last Post
Replies
3
Views
980
  • Last Post
Replies
2
Views
1K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
4
Views
819
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
900
  • Last Post
2
Replies
28
Views
4K
Top