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Help me separate this equation

  1. Apr 13, 2007 #1

    ssb

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

    Consider the following differential equation
    y' = t*ln(y^(2t))+t^2

    There is one constant solution to this problem. Find it
    Basically I have to get y' and all other y on the left, and leave all the t's on the right.
    2. Relevant equations



    3. The attempt at a solution

    Its just algebra but I cant remember all the rules with e and ln.
     
    Last edited: Apr 13, 2007
  2. jcsd
  3. Apr 13, 2007 #2
    remember:

    ln(a^b) = b*ln(a)

    what does that allow you to do?
     
  4. Apr 13, 2007 #3

    Dick

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    Luckily, you don't have to solve this equation to answer the question. If y is a constant solution what can you say about y'?
     
  5. Apr 13, 2007 #4

    ssb

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    ChaoticLlama Thanks for that tip. I had forgotten that one. I went on using your tip and I came across this when I separated the equations:

    1/ln(y) dy = 2t^2 + t dt

    First off im not sure how to integrate 1/ln(y)

    -----------------------------

    Dick, if y' has a constant solution, then y at a given point has a constant solution? are you suggesting that I dont have to solve one side for y and I can just plug in the numbers or something? confused

    Thanks for the help guys!
     
  6. Apr 13, 2007 #5

    Dick

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    Well, you didn't separate the equation correctly. So I wouldn't worry about integrating 1/ln(y). I was trying to get you to realize that if y is a constant solution then y'=0. The trouble with this is that the equation as written doesn't seem to have any constant solutions. As you sure you transcribed it correctly??
     
  7. Apr 13, 2007 #6
    Are you sure that you typed out the DE correctly? Is it

    [tex]y^\prime = t\ln \left(y^{2t}\right) + t, \ \mbox{or } \ y^\prime = t(\ln \left(y^{2t}\right) + t)?[/tex]

    The first one (which is what's in your first post) does not have a constant solution at all. The second one does.

    With regard to Dick's suggestion, think about it: If y is constant, then what's dy/dt?
     
  8. Apr 13, 2007 #7

    ssb

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    jesus your right. im such an idiot. i even double checked it and triple checked it too..... the last T has is squared

    y' = t*ln(y^(2t))+t^2

    I feel so stupid
     
  9. Apr 13, 2007 #8
    Okay, that's the second one I posted up there. So use Dick's suggestion and solve the resulting eq. :smile:
     
  10. Apr 13, 2007 #9

    Dick

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    Happens to the best of us. Now it has a constant solution. Can you find it?
     
  11. Apr 13, 2007 #10

    ssb

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    [tex]\ y^\prime = t \ln \left(y^{2t}\right) + t^2[/tex]

    [tex]\ y^\prime = t * (2t) \ln \left(y) +t^2[/tex]

    [tex]\ 1/\ln \left(y) y^\prime = t (2t) + t^2 [/tex]

    [tex]\ 1/\ln \left(y) y^\prime = 3t^2 [/tex]

    [tex]\ \int 1 / ln (y) (y^\prime) = \int 3t^2 [/tex]

    [tex]\ \int 1 / ln (y) (dy) = \int 3t^2 dt[/tex]

    [tex]\ ? = t^3 + C [/tex]


    With your clues that you were giving me on the fact that y was constant, i thought that you meant that

    [tex]\ 0 = t^3 + C [/tex]

    but I think im going the wrong way with this
     
    Last edited: Apr 13, 2007
  12. Apr 13, 2007 #11
    I'll repeat again: If y is constant, then what's y' = dy/dt (remember, intuitively dy/dt is the rate of change of y with respect to t)?
     
  13. Apr 13, 2007 #12

    ssb

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    It would be zero. Give me a minute to think about this one.... i know the significance should probably be smacking me in the face right about now. All I can think is that the derivative of a constant is zero so part of my equation should equal zero?

    In post 10 im trying to show my steps because im just not getting it and im sorry. Calculus is really killing me this quarter.
     
  14. Apr 13, 2007 #13

    HallsofIvy

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    Since your equation is y'= f(x) the whole equation should be 0! If y(t) is a constant function then
    y'(t)= t*ln(y^(2t))+t^2= 0. Now solve for y (remember that it should NOT depend on t because it is a constant). (Of course, the fact that [itex]ln(y^{2t})= 2t ln(y)[/itex] is useful!)
     
  15. Apr 13, 2007 #14

    Dick

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    You can't separate the variables in that equation unless you make an algebraic mistake. Which you did. Then other mistakes ensue. I don't know why you won't believe the advice you've been given repeatedly that you don't have to solve a differential equation.
     
  16. Apr 14, 2007 #15

    ssb

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    Im sorry. I just dont understand the advice i guess. I have yet to successfully solve one of these problems for my class. Im used to single variable calculus and this is the first time ive ever seen 2 variables in a math problem such as this. its the first week of class and the teacher expects us to already know how to do this kind of stuff. Last quarter we didnt cover these in any of our topics.

    Maybe do you know of a good website that shows example problems worked out step by step so I can see whats expected to be done? ive searched google and wikipedia and cannot find any good examples.

    All i know is that hes looking for a "Constant solution" to this problem. I dont know if the solution will contain just numbers, y, t, or a combination of the 3. To be quite honest im not even sure what form to put such an answer in. sorry.

    The equation is
    [tex]y'=t\,\ln(y^{2t})+t^2[/tex]
    and the question states that it is separable. Can someone help me separate it and let me see if I can get it from there? Thanks
     
    Last edited: Apr 14, 2007
  17. Apr 14, 2007 #16
    What does constant mean? A constant solution should contain only constants.

    It is separable, but since you are looking for a constant solution, as numerous others have pointed out, this is not the way to go about this problem.
     
  18. Apr 15, 2007 #17

    HallsofIvy

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    I told you before, if y is a constant then its derivative is 0 so you have y'(t)= t*ln(y^(2t))+t^2= 0. Solve the equation t*ln(y^(2t))+t^2= 0 for y!
     
  19. Apr 15, 2007 #18

    HallsofIvy

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    I told you before: if y is a CONSTANT solution, then [itex]y'(t)= t ln(y^{2t})+t^2= 2t^2 ln(y)+ t^2= 0[/itex]. Solve that for y.
     
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