text: 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
explanation: This document is a discussion thread from a forum where a user, mikehibbert, is seeking help with a differential equation problem. The problem is to solve the differential equation \( \frac{dy}{dx} = \frac{2x}{y+1} \) using separation of variables. The user initially gets stuck and posts their attempt, which leads to a series of responses from other forum members, including jeffreydk and tiny-tim, who provide guidance and hints. The discussion continues with the user attempting to solve the equation and receiving further corrections and advice.
  
  The document contains a question and multiple responses, including hints and partial solutions, but does not contain a complete, final answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: This document is a discussion thread from a forum where a user, mikehibbert, is seeking help with a differential equation problem. The problem is to solve the differential equation \( \frac{dy}{dx} = \frac{2x}{y+1} \) using separation of variables. The user initially gets stuck and posts their attempt, which leads to a series of responses from other forum members, including jeffreydk and tiny-tim, who provide guidance and hints. The discussion continues with the user attempting to solve the equation and receiving further corrections and advice.
  
  The document contains a question and multiple responses, including hints and partial solutions, but does not contain a complete, final answer.
  
  Label: multiple-question-and-answer