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Losing a solution of a 1st order ODE?

  1. Mar 10, 2005 #1
    There is an example in a book regarding DEs which I do not understand. Solve the IVP

    [tex] y'=y^2-4, y(0)=-2 [/tex]where t is the independent variable

    We first solve by separation of variables to arrive at the 1-parameter solution.

    [tex] -\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c[/tex]

    Simplifying and expressing the solution explicitly, we find that,

    [tex] y=2\frac{1+ce^{4t}}{1-ce^{4t}} [/tex]

    Taking the initial condition,

    [tex] -2=2\frac{1+c}{1-c} [/tex] which simplifies to,


    They said that the solution is wrong because:

    we can express the DE as, [tex] y'=(y+2)(y-2) [/tex] and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?
    Last edited: Mar 10, 2005
  2. jcsd
  3. Mar 10, 2005 #2


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    Your initial condition y(0)=2 seems to imply that c = 0, not -1=1?
    You can preclude y=-2 simply by noticing that it fails to satisfy the initial condition.
  4. Mar 10, 2005 #3
    sorry, my initial condition is [tex] y(0)=-2[/tex]. Do we ignore [tex]y=-2[/tex] because firstly, it doesn't tell us the value of c, and it gives us the illogical [tex]-1=1[/tex]?

    Also, do we use the other alternative [tex]y=2[/tex] because it is the other "alternative" in [tex]y'=(y+2)(y-2)[/tex]??? If it is so, why?
    Last edited: Mar 10, 2005
  5. Mar 10, 2005 #4


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    The "separable", non-constant solutions involve a division of both sides of the equation with the expression [tex]y^{2}-4[/tex]
    But, since you can't divide by zero, you have implicitly assumed that [tex]y\neq\pm2[/tex]

    In addition to the non-constant solutions, you've got the constant solutions [tex]y_{1}(t)=2,y_{2}(t)=-2[/tex]
    Last edited: Mar 10, 2005
  6. Mar 10, 2005 #5


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    [tex] y(0)=-2 [/tex] is not a valid initial condition.When separating variables YOU ASUMED [tex] y\neq 2;y\neq -2 [/tex] Trying to impose the initial condition to

    [tex] \frac{y-2}{y+2}=Ce^{4t} [/tex]


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