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Homework Help: First order, non linear ode problem

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Finding the general solution of y'= (2y^2)/e^x
    and the particular given that y(0) = 0


    3. The attempt at a solution

    I seperate variables etc. to get y = -1/(2c-2e^(-x)) where c is the constant. The problem then is obtaining the particular solution. If I substitute y and x for 0 i am left with 0= -1/(2c-2) which is unsolvable. what am i doing wrong ?
     
  2. jcsd
  3. May 11, 2010 #2

    Dick

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    Your equation is singular at y=0. There's another obvious solution you are missing.
     
  4. May 11, 2010 #3
    Ok.. I also get this as a solution y = -e^x/(ce^x-2) but this is also singular at y = 0.. i cant see the other obvious solution.
     
  5. May 11, 2010 #4

    Dick

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    That's the same as your first solution. If an ode is singular, it may have more than one solution. So there may be another one besides the one you found. How about y is identically equal to zero?
     
  6. May 11, 2010 #5
    yeah, i dont understand what you mean when you say that its singular. Does that mean the particular solution does not exist?
     
  7. May 11, 2010 #6
    The way I see it, you're confusing first-order ODEs with second-order ODEs.

    In a first-order ODE, there is no particular solution, but instead a general solution, which is just a family of functions for different "c"s, that is, constants.

    Once you're given an IVP (Initial Value Problem), then you'll find a value for the constant "c", which leads to a unique solution to your first-order ODE.

    Don't use the term "particular solution" for first-order ODEs, use "unique solution" instead, that's because in second-order ODEs, you generally have two solutions called "complementary solution" and "particular solution", so be careful with the nomenclatures, as that can make a big confusion.

    For example, your first-order ODE is

    [tex]y' = \frac{2y^2}{e^x}[/tex]

    You said that you separated the variables, so that leads us to

    [tex]\frac{dy}{y^2} = \frac{2}{e^x}dx[/tex]

    Which can also be rewritten as

    [tex]\frac{dy}{y^2} = 2e^{-x} dx[/tex]

    So what do you do now? You integrate both sides

    [tex]\int \frac{dy}{y^2} = \int 2e^{-x} dx[/tex]

    [tex]\frac{-1}{y} = -2e^{-x} + c [/tex]

    ATTENTION: Always remember your constants, after the integration process.

    So I believe you can take it from here. Also, don't forget that the problem asks for a specific solution, that is, a unique solution, and not the general solution, so that means you'll have to solve for "C", and that can be done by using your initial value problem, which was given.
     
    Last edited: May 11, 2010
  8. May 11, 2010 #7
    thanks for the detailed explanation. So I need to find the unique solution, and you have taken me to the part were i fail to find it. This is my blockade. The IVP is when y(0) = 0 and when i try to solve for "c" it is not possible. Uniqueness fails. I think this is what Dick was trying to tell me with y being identically equal to zero. Although I dont see exactly why that is so.
     
  9. May 11, 2010 #8
    When you separate variables, you are assuming y is not zero. Otherwise, you couldn't divide by y.
     
  10. May 11, 2010 #9

    Dick

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    I think I'm using 'singular' in the wrong way, but the solution y(x)=0 for all x solves your equation. In that case you can't divide both sides of your equation by y^2 and use separation of variables. Because y=0!
     
  11. May 11, 2010 #10
    Hmm, that's true, it can't be done by separation of variables then.

    But is there a solution then?
     
  12. May 11, 2010 #11
    Ok. I think I sort of got the hang of what is happening in this problem. Yeah, there is no other possibility except y(x)= 0 . It is the only solution that would work for the given IVP. Thanks,
     
  13. May 11, 2010 #12

    Dick

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    y(x)=0. Isn't that a solution?
     
  14. May 11, 2010 #13
    trick problem :mad:
     
  15. May 11, 2010 #14

    Mark44

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    And I think you are confusing homogeneous differential equations with their nonhomogeneous counterparts.

    Let's look at a simple example.
    Homogeneous: y' - 2y = 0
    Nonhomogeneous: y' - 2y = x

    For the homogeneous equation, it's easy to see that the general solution is y = Ae2x. If this had been an initial value problem we could use the initial condition to solve for A.

    For the nonhomogeneous equation, a particular solution is yp = -x. As before, the complementary solution is yc = Ae2x, so the general solution is y = yc + yp = Ae2x - x. Given an initial condition we could solve for A, arriving at a unique solution to the nonhomogeneous problem.
     
  16. May 11, 2010 #15

    You're absolutely correct Mark. What I meant was in fact homogeneous and nonhomogeneous equations just as you said, and somehow mislabelled it as first-orders and second-orders. Thanks for correcting me up.
     
  17. May 11, 2010 #16

    vela

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    Actually, what particular solution means depends on the context. Some texts use the term to mean the solution to the IVP, as opposed to the general solution, and it's in this sense that the OP has used it. I don't see it as any sign of confusion on his or her part.
     
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