1. Oct 20, 2008

### mikehibbert

1. The problem statement, all variables and given/known data

dy/dx = 2x / (y+1)

2. Relevant equations

n/a

3. The attempt at a solution

I have to solve this - and i know it's separation of variables, but I always get y on both sides when I try to solve and end up with:

y = 2(x^2) / (y+1)

which is clearly wrong.

Where am I messing up? :S

2. Oct 20, 2008

### jeffreydk

Sorry the latex doesn't seem to be working, but since it's seperable, go...

integral[ (y+1) dy] = integral[ 2x dx]

and when you integrate that you should be on your way.

3. Oct 20, 2008

### mikehibbert

that's what i did - but that came out as the solution i gave in my original post - which isn't right is it?

4. Oct 20, 2008

### tiny-tim

Hi mikehibbert!
No, it doesn't … try again!

5. Oct 20, 2008

### mikehibbert

integrating (y+1) wrt y gives: (y^2)/2 + y

and integrating 2x wrt x gives: x^2

giving (y^2)/2 + y = x^2

and rearranged gives y = 2x / (y+1)

i still get it no matter how many times i think of it :(

arrgghhh

6. Oct 20, 2008

### tiny-tim

I assume you mean y = 2x2 / (y+1) …

but shouldn't there be a 2y in there somewhere?

and what about a constant of integration?

7. Oct 20, 2008

### mikehibbert

yes i did mean y = 2(x^2) / (y+1) sorry.

and there should be a +c on the end, granted.

but my solution is still wrong isn't it?

you can't have a y on both sides?

8. Oct 20, 2008

### mikehibbert

oh and i can't see where a 2y would come from??

the LHS integral is (y^2) / 2 + y

which factorizes to 0.5y(y+1) ???

then re-arranges to what i said earlier.

9. Oct 20, 2008

### tiny-tim

Nope … 0.5y(y+2)

10. Oct 20, 2008

### mikehibbert

grrr i'm literally an idiot...

11. Oct 20, 2008

### mikehibbert

i swear i'm going to kill the person than wrote this question...

now i'm getting y = 2(x^2) + x^2 + c

which doesnt really look right to me :S

12. Oct 20, 2008

### tiny-tim

D'oh!

Get some sleep … :zzz:

13. Oct 20, 2008

### mikehibbert

i'm gunna have to.
gotta be handed in at 10am tomo grrr.
i give in :P
the easiest question on the sheet has beaten me!

14. Oct 20, 2008

### tiny-tim

It'll all be clear at breakfast!

15. Oct 20, 2008

### jeffreydk

Just rewrite it as y^2+y-2x^2-C=0, and you want to factor it. So use the quadratic formula and you'll be able to get the two solutions for the differential equation.

16. Oct 21, 2008

### HallsofIvy

Staff Emeritus
I don't think that is the right person to kill!

You have, correctly, (1/2)y2+ y= x2+ C and you want to solve for y. That is a quadratic equation:
y2+ y- (x2+ C)= 0. Use the quadratic formula.

17. Oct 21, 2008

### tiny-tim

Hey guys!

Why do you keep writing the wrong equations?

18. Oct 21, 2008

### Geekchick

Hi!
I'd like to venture a guess (I'm currently in the middle of calculus 1 and haven't really done any differential equations)

For the integral of (y+1) I have [(y^2)/2] + y

For the integral of 2x I have x^2

then I set them equal and solved for x to get

x = the +-squareroot of {[y^2+2y]/2} I thought it looked nicer to give them a common denominator.

Anyway since originator of the post has already turned in the assignment I didn't see what it would hurt to post a possible answer, but anyway could someone let me know if I am anywhere close or if not post the right answer.

Thank!

19. Oct 21, 2008

### tiny-tim

Hi Geekchick!

i] You must include a constant of integration.

ii] Don't use √s … as HallsofIvy said, this is a quadratic equation … so leave it as such … it tells you what shape the curve is.

20. Oct 21, 2008

### Geekchick

alright...Best I can come up with is y=-1+-square root of (2x^2+C)

Last edited: Oct 21, 2008